Hakchin Kim
June 24, 2021
Contents
1 Reference Sites 8
2 Concepts 10
2.1 Logic Initials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 log . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 12
3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4 17
4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.1.3 i , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2.2 , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
5 21
5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5.1.1 이의 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Contents 3
5.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
5.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
5.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
6 28
6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
6.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
6.1.2 이의 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
6.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6.2.3 건과 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6.2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
7 38
7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
7.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
7.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
7.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
7.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
8 48
8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
8.1.1 곱과 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
8.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
8.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
8.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
8.4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
9 51
9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
9.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
9.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
9.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
9.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
9.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
9.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
9.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
9.3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Contents 4
9.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
10 57
10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
10.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
10.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
10.2.3 이의 . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
10.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
10.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
10.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
10.5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
10.5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . 64
11 65
11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
11.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
11.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
11.7 이의 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
11.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
11.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
12 67
12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
12.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
12.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
12.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
12.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
12.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
12.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
12.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
12.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
13 69
13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
13.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
13.1.2 하학 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
13.1.3 속성 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
13.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
13.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
13.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
13.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
Contents 5
13.2.4 , . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
13.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
13.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
13.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
13.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
14 79
14.1 정적 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
14.1.1 Reference Sites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
14.2 정적 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
14.3 정적 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
15 82
15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
15.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
15.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
15.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
15.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
15.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.3.3 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
15.4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
15.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
15.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
15.6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
15.6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
16 91
16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
16.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
16.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
16.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
16.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
16.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
16.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
16.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
17 93
17.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
17.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
17.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
17.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
17.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
17.2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
Contents 6
17.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
17.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
17.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . 101
17.3.3 율의 율의 . . . . . . . . . . . . . . . . . . . . . . . . . . 104
18 108
18.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
18.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
18.3 Hilbert’s axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
18.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
19 111
20 112
20.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
20.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
20.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
20.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
20.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
20.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 116
20.2.2 . . . . . . . . . . . . . . . . . 116
20.2.3 . . . . . . . . . . . . . . . . . . . . . . . . 116
21 공간 118
21.0.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
21.0.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
22 공간 120
22.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
22.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
22.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
22.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
22.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
22.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
22.6 , orthograph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
22.7 공간 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
22.7.1 Reference Sites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
23 133
23.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
23.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
23.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
필함 해할
부분니다. 다는 해할
적절 개가 니다. Definition,
Theorem 니다.
니다.
니다.
1. .
2. .
.
다는 .
, .
. 소설 작을, 세서
인의 . .
3. It’s better when it’s simple
1 Reference Sites
List of mathematical symbols
Intro Probability Theory
계과
link-
link- ?
link-Mathematical proof
link-List of logic symbols
link-Symbol Contradiction
List of mathematical abbreviations
Contradiction
외우
임의의 .
Introduction to Higher Mathematics -
(m,n) = Gcd(n, Mod(m, n))
Euclidean Algorithm -
8
godingMath- - .
하학
1.3
Figure 1.1: A caption, explaining that this is a test.
1 Reference Sites 9
Figure 1.2: absolute size test for images
Figure 1.3: relative size test for images
Figure 1.4: relative size test for images
2 Concepts
2.1 Logic Initials
S cond : a sufficient condition
N cond : a necessary condition; a requirement
2.2
U : Universal set
Z : ganze Zahl(Integer)
Q : Quotient(Rational number)
P = R \Q : Irrational number
E : Interval [0,1]
I : An Interval
C : Complex number
N : Natural number
W = N {0} : Whole number
R : Real number
F : Set of functions
S : Solutions of equations, inequalities, etc.
¨
I(R) : Set of functions which are strictly increasing on R.
¨
i(R) : Set of functions which are monotonically increasing on R.
PIN : Purely Imaginary Number
PN : Prime Numbers
2.3 log
Definition 2.3.1 (log). a
m
= N m = log
a
N
At this time, we call m a N
N log
a
N
2.4 11
2.4
2 : 2
3 : 3
4 : 4 (100 4 )
5 : 5
6 : 2 3 ( 3 )
8 : 3 8 (1000 8 )
9 : 9
3
3.1
3.1.1
Definition 3.1.1 ().
1. 자의 .
2. 자의 ,
부분 .
Definition 3.1.2 ().
1. , 각각
.
2. .
3. 함하 , .
Definition 3.1.3 ( ).
1. : 각각
2. : 부분
3. :
4. :
5. :
6. :
Figure 3.1:
Definition 3.1.4 ( ).
1. :
3.1 13
2. :
Example 3.1.1 ( 2xy + 3x
2
y
2
+ 4x + 1).
x 3x
2
+ (2y + 4)x y
2
+ 1
x y
2
+ 1 + (2y + 4)x + 3x
2
Definition 3.1.5 ( ).
1. : A, B A+B ,
.
2. : A, B AB A B B
.
Note ( ).
.
Figure 3.2:
1. .
2. .
3. .
Theorem 3.1.1 ( ).
A, B, C
1. : A+B=B+A
2. : (A+B)+C=A+(B+C) L99 A+B+C .
3.1.2
Definition 3.1.6 ( ).
A, B AB 배법 ,
.
Theorem 3.1.2 ().
a, b 실수, m, n , .
1.
2.
3.1 14
1. (a + b)
2
= a
2
+ 2ab + b
2
(a b)
2
= a
2
2ab + b
2
2. (a + b)(a b) = a
2
b
2
3. (ax + b)(cx + d) = acx
2
+ (ad + bc)x + bd
4. (x + a)(x + b)(x + c) = x
3
+ (a + b + c)x
2
+ (ab + bc + ca)x + abc
5. (ax + b)(cx + d)(ex + f ) = acex
3
+ (acf + ceb + ead)x
2
+ (adf + cfb + ebd)x + bdf
6. (a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ca
7. (a + b)
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
(a b)
3
= a
3
3a
2
b + 3ab
2
b
3
8. (a + b)(a
2
ab + b
2
) = a
3
+ b
3
(a b)(a
2
+ ab + b
2
) = a
3
b
3
9. (a
2
+ ab + b
2
)(a
2
ab + b
2
) = a
4
+ a
2
b
2
+ b
4
10. (a + b + c)(a
2
+ b
2
+ c
2
ab bc ca) = a
3
+ b
3
+ c
3
3abc
11. (a + b + c)(ab + bc + ca) = (a + b)(b + c)(c + a) + abc
12. (a b)(b c)(c a) = ab
2
+ bc
2
+ ca
2
(a
2
b + b
2
c + c
2
a)
13. (a + b)
n
=
P
n
k=0
C(n, k)a
nk
b
k
=
P
n
k=0
C(n, k)a
k
b
nk
14. (a +
1
b
)(b +
1
a
) = ab + 2 +
1
ab
15.
1
ab
=
1
b a
(
1
a
1
b
) (where a 6= b)
16.
1
abc
=
1
c a
(
1
ab
1
bc
) (where a 6= c)
3.1.3
1. a
3
+ b
3
= (a + b)
3
3ab(a + b) = (a + b)(a
2
ab + b
2
)
a
3
b
3
= (a b)
3
+ 3ab(a b) = (a b)(a
2
+ ab + b
2
)
2. (a + b)
2
+ (b + c)
2
+ (c + a)
2
= 2(a
2
+ b
2
+ c
2
) + 2(ab + bc + ca)
a
2
+ b
2
+ c
2
+ (ab + bc + ca) =
1
2
((a + b)
2
+ (b + c)
2
+ (c + a)
2
)
3. (a b)
2
+ (b c)
2
+ (c a)
2
= 2(a
2
+ b
2
+ c
2
) 2(ab + bc + ca)
a
2
+ b
2
+ c
2
(ab + bc + ca) =
1
2
((a b)
2
+ (b c)
2
+ (c a)
2
)
4. a
n
b
n
= (a b)(
P
n
k=1
a
nk
b
k1
) = (a b)(
P
n
k=1
a
k1
b
nk
)
a
n
+ b
n
= (a + b)(
P
n
k=1
a
nk
b
k1
(1)
k1
) = (a b)(
P
n
k=1
a
k1
b
nk
(1)
nk
)
3.2 15
x
n
1 = (x 1)(
P
n
k=1
x
nk
)
x
n
+ 1 = (x + 1)(
P
n
k=1
x
nk
(1)
k1
)
3.1.4
Definition 3.1.7 ( ). A, B (B 6= 0) A÷B
A = BQ + R (where (R) ¡ (B))
Q R . Q , R .
R = 0 A B A=BQ .
3.2
Definition 3.2.1 ().
(=) 음을 .
자의 , .
자의 .
Definition 3.2.2 (()).
지지
Definition 3.2.3 ().
: 방법
: 방법
3.2.1
Example 3.2.1 ( Tip).
50
100
2499 P (x) = (x
2
1)Q(x) + ax + b .
8
10
7 , 2
100
31 P (x) = (x + 1)
n
= xQ(x) + 1
.
2
1111
17 9
Theorem 3.2.1 (). ,
. .
f(x) x a R
R = f(a)
3.2 16
f(x) ax + b R
R = f(
b
a
)
Theorem 3.2.2 (Hakchin).
f(x) = A(x)B(x) + C(x) Mod(f(x), A(x)) = Mod(C(x), A(x))
Theorem 3.2.3 (). f(x) x a
fa. .
1. f(x) x α f(α) = 0.
2. f(α) = 0 f(x) x α .
3.2.2
4
4.1
4.1.1
Definition 4.1.1 ( i). 1 , i .
i
2
= 1, i =
1
Definition 4.1.2 (). 실수 a, b i a + bi
, a 실수부분, b 부분 . a + bi
.
실수 a + bi b = 0
a + bi b 6= 0
a + bi a = 0, b 6= 0
Note ( ).
4.1.2
Definition 4.1.3 ((Complex conjugate)).
a + bi (a, b R) 부분 a bi a + bi ,
a + bi .
a + bi = a bi
Theorem 4.1.1 ( ).
1.
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) (c + di) = (a c) + (b d)i
2.
(a + bi)(c + di) = (ac bd) + (ad + bc)i
3.
(a + bi)
(c + di)
=
(a + bi)(c di)
(c + di)(c di)
=
(ac + bd) + (ad + bc)i
(c
2
+ d
2
)
Theorem 4.1.2 ( ). z
1
, z
2
z
1
, z
2
4.1 18
1. (z
1
) = z
1
2. z
1
+ z
2
= z
1
+ z
2
, z
1
z
2
= z
1
z
2
3. z
1
· z
2
= z
1
· z
2
, (
z
1
z
2
) =
z
1
z
2
(where z
2
6= 0)
4. z + z R, z · z R
5. z
n
= (z)
n
6. z z = 0 z R
7. z + z = 0 (z PIN z = 0)
8. z
2
0 z R
9. z
2
< 0 z PIN
10. z
2
R z R z PIN
11. (z = ±a + ai z = a ± ai) (z
2
PIN z = 0)????
4.1.3 i ,
Theorem 4.1.3. k [0, ) Z
i
4k
= 1
i
4k+1
= i
i
4k+2
= 1
i
4k+3
= i
Theorem 4.1.4 ( ).
a
b
=
r
a
b
(a > 0 b < 0) (a = 0 b 6= 0) (4.1)
a
b =
ab (a < 0 b < 0) (a = 0) (b = 0) (4.2)
Proof. ,
(a > 0 b < 0)
a
b
=
a
bi
=
ai
bi
2
=
ai
b
=
r
a
b
i =
r
a
b
=
r
a
b
(a < 0 b < 0)
a
b =
ai
bi =
p
(a)(b)i
2
=
ab
Corollary 4.1.4.1 (음의 실수 ).
1. a > 0
a =
ai
2. a > 0 Sqrt(a) = ±
ai J (
ai)
2
= a · i
2
= a, (
ai)
2
= a · i
2
= a
4.2 19
3. a < 0
a =
ai
4.1.4
4.1.5
Theorem 4.1.5 ( ).
ax
2
+ bx + c = 0 (a = 0) α, β
α + β =
b
a
, αβ =
c
a
Theorem 4.1.6 ( ).
ax
2
+ bx + c = 0 (a = 0)
1. a, b, c , p + q
m p q
m. (where p, q Q, q 6=
0,
m I)
2. a, b, c 실수 , p + qi p qi. (where p, q R, q 6= 0, i =
1)
Remark ( ). x
2
+
3x 1 +
3 = 0 (x + 1)(x 1 +
3) = 0
4.2
4.2.1
4.2.2 ,
4.3
4.3.1
Note ( ).
, Apollonios’ Theorem,
4.3.2
2
Note ( ).
ax
4
+ bx
2
+ c = 0
1. x
2
= t .
4.4 20
2. A
2
B
2
.
Figure 4.1: 해하
4.4
4.4.1
함한
4.4.2
실수 .
5
5.1
5.1.1 이의
이의
5.1.2
Note (Mid Point).
Mid(x
1
, x
2
) = Mid(0, x
1
+ x
2
)
Theorem 5.1.1 (Stewart’s Theorem).
Tikz .
Figure 5.1: Stewart’s Theorem
mb
2
+ nc
2
= (m + n)(mn + d
2
) = a(mn + d
2
)
Proof. cosine 2 .
5.2
Theorem 5.2.1 (Fermat point). P
Proof.
르마
5.2 22
르마
르마
르마
르마
5.2.1
Definition 5.2.1 ( ). sfjsldajflsdaj
Theorem 5.2.2 ( ). sfjsldajflsdaj
Theorem 5.2.3 ( 이의 ).
1. (x
1
, y
1
) ax+by+c=0 이의 d
d =
|ax
1
+ by
1
+ c|
a
2
+ b
2
2. ax+by+c=0 이의 d
d =
|c|
a
2
+ b
2
5.2.2
Definition 5.2.2 (수심, an orthocenter).
수심 니다. 수심 니다.
Theorem 5.2.4 (수심).
5.2 23
Figure 5.2: 수심
Note ( ).
이의
Theorem 5.2.5 ( ).
5.3 24
Figure 5.3: triangle
5.3
Definition 5.3.1 (). 위의 정점 .
정점 , 리를 .
5.3.1
Theorem 5.3.1 ( ). C(a,b) r
(x a)
2
+ (y b)
2
= r
2
Theorem 5.3.2 ( ).
Ax
2
+ By
2
+ Cx + Dy + E = 0 where (A = B 6= 0) (C
2
+ D
2
4AE > 0)
(
C
2A
,
D
2A
)
C
2
+ D
2
4AE
2A
Theorem 5.3.3 ( ).
(x
2
+ y
2
+ Ax + By + C) + k(x
2
+ y
2
+ A
0
x + B
0
y + C
0
) = 0 where k R {−1}
Theorem 5.3.4 ( ).
(x
2
+ y
2
+ Ax + By + C) (x
2
+ y
2
+ A
0
x + B
0
y + C
0
) = 0
Theorem 5.3.5 ( - 1).
x
2
+ y
2
= r
2
, m
y = mx ±r
p
m
2
+ 1
5.4 25
Theorem 5.3.6 ( -접점 2).
x
2
+ y
2
= r
2
위의 P (x
1
, y
1
)
xx
1
+ yy
1
= r
2
Theorem 5.3.7 ( -접점 3).
(x a)
2
+ (y b)
2
= r
2
위의 P (x
1
, y
1
)
(x a)(x
1
a) + (y b)(y
1
b) = r
2
Theorem 5.3.8 (circle of Apollonios, ).
A, B AP : P B = m : n(m, n > 0) P AB m:n
. (Apollonios)
.
Figure 5.4: circle of Apollonios
5.3.2
5.4
5.4.1
5.4.2
Note ( ).
5.4 26
Figure 5.5:
Note ( y = mx + n ).
Condition 1 : A(x
1
, y
1
), B(x
2
, y
2
), l : y = mx + n
Condition 2 : B = ST (A, l)
Conclusion (
y
2
y
1
x
2
x
1
· m = 1) (
y
1
+ y
2
2
= m ·
x
1
+ x
2
2
+ n)
Note ( ±1 ).
1. y = x + n
5.4 27
(x, y) (y n, x + n)
f(x, y) = 0 f (y n, x + n) = 0
2. y = x + n
(x, y) (y + n, x + n)
f(x, y) = 0 f (y + n, x + n) = 0
6
6.1
6.1.1
Definition 6.1.1 ().
Definition 6.1.2 (). ainA
Definition 6.1.3 ( 방법).
1.
2.
3. : 안에
방법
Definition 6.1.4 ().
Definition 6.1.5 ().
Definition 6.1.6 (). , .
Definition 6.1.7 ( ). A , A
n(A) .
6.1.2 이의
Definition 6.1.8 (부분). A B x A x B
Definition 6.1.9 (부분). A B 부분 A B A 6= B
Theorem 6.1.1 (부분 ).
1. n(A) = k A 부분 2
k
2. A = {a
1
, a
2
, a
3
, ..., a
n
} k 부분 2
nk
(, n k)
3. A = {a
1
, a
2
, a
3
, ..., a
n
} k 부분
2
nk
(, n k)
6.1.3
Definition 6.1.10 (). A B = x|x A x B
Definition 6.1.11 (). A B = x|x A x B
6.2 29
Definition 6.1.12 (). A B =
Definition 6.1.13 (). 부분 ,
, U .
Definition 6.1.14 ().
Definition 6.1.15 ().
Theorem 6.1.2 ( ). 1.
2.
3. 배법
Theorem 6.1.3 ( ).
1. (A B)
c
= A
c
B
c
2. (A B)
c
= A
c
B
c
Definition 6.1.16 ().
A 4 B = (A B) (A B)
c
= (A B) (B A)
Definition 6.1.17 ().
A 5 B = (A B)
c
(A B)
Theorem 6.1.4 ( ).
1. .
2. A 4 B = (A 5 B)
c
= A
c
4 B
c
3. A 5 B = (A 4 B)
c
= A
c
5 B
c
6.2
6.2.1
6.2.2
6.2.3 건과
6.2.4
Definition 6.2.1 (). 실수
Definition 6.2.2 ((A), an Arithmetical mean).
Definition 6.2.3 ((G), a Geometric mean). n n
Definition 6.2.4 ((H), a Harmonic mean).
Theorem 6.2.1. 음이 .
A G H
6.2 30
Example 6.2.1. i.e. a, b R
+
,
(a + b)
2
ab
2ab
a + b
Theorem 6.2.2.
(ax + by)
2
(a
2
+ b
2
)(x
2
+ y
2
) (6.1)
(ax + by + cz)
2
(a
2
+ b
2
+ c
2
)(x
2
+ y
2
+ z
2
) (6.2)
(a
1
b
1
+ a
2
b
2
+ ... + a
n
b
n
)
2
(a
2
1
+ a
2
2
+ ... + a
2
n
)(b
2
1
+ b
2
2
+ ... + b
2
n
) (6.3)
b
1
a
1
=
b
2
a
2
= ... =
b
n
a
n
Question 6.2.1. a, b, c R, ax
2
+ bx + c 0 .
Solution.
ax
2
+ bx + c
= a(x
2
+
b
a
x +
c
a
)
= a{x
2
+
b
a
x + (
b
2a
)
2
+
c
a
(
b
2a
)
2
}
= a{(x +
b
2a
)
2
+
c
a
b
2
4a
2
}
= a(x +
b
2a
)
2
+ a(
4ac b
2
4a
2
)
= a(x +
b
2a
)
2
+
4ac b
2
4a
a > 0 4ac b
2
0
Question 6.2.2. a, b R, .
.
a
2
+ b
2
ab
Hint.
a b > 0 a > b
Solution.
a
2
+ b
2
ab
= a
2
ab + (
b
2
)
2
(
b
2
)
2
+ b
2
= (a
b
2
)
2
+
3
4
b
2
0
a
2
+ b
2
ab 0 a
2
+ b
2
ab
Question 6.2.3. . .
a, b R, |a| + |b| |a + b|
6.2 31
Solution.
a
2
= |a|
2
.
(|a| + |b|)
2
|a + b|
2
= a
2
+ 2|a||b| + b
2
(a
2
+ 2ab + b
2
)
= a
2
+ 2|ab| + b
2
a
2
2ab b
2
= 2|ab| 2ab
= 2(|ab| ab) 0
|a| + |b| |a + b|
, , |ab| = ab .
(a = 0 b = 0) (a > 0 b > 0) (a < 0 b < 0)
Question 6.2.4. a > 0, b > 0, ab = 9 , a + b .
Solution.
a + b 2
ab
a + b 2 ·
9
a + b 6
6
Question 6.2.5. x > 0 , x +
1
x
.
Hint.
A G, i.e. a, b R
+
,
(a + b)
2
ab
Solution.
A G,
x +
1
x
2
r
x ·
1
x
= 2
2
Question 6.2.6. x > 3 , x + 2 +
4
x 3
.
Solution.
x + 2 +
4
x 3
x 3 +
4
x 3
+ 5
x 3 +
4
x 3
+ 5 2
s
(x 3) ·
4
(x 3)
+ 5 = 4 + 5
9
6.2 32
Question 6.2.7. x < 0 , x +
4
x
.
Solution.
x > 0,
x +
4
x
2
s
(x) ·
4
(x)
= 4
x +
4
x
4
4
Question 6.2.8. x > 0 ,
2x
x
2
+ 3x + 4
.
Solution.
2x
x
2
+ 3x + 4
x ,
2
x + 3 +
4
x
x + 3 +
4
x
2
r
x ·
4
x
+ 3 = 7
2
x + 3 +
4
x
2
7
2
7
Question 6.2.9. a > 0, b > 0 , (a + b)(
1
a
+
9
b
) .
Hint.
A G, i.e. a, b R
+
,
(a + b)
2
ab
Solution.
(a + b)(
1
a
+
9
b
)
= a ·
1
a
+ a ·
9
b
+ b ·
1
a
+ b ·
9
b
= 1 + 9 ·
a
b
+
b
a
+ 9
2
r
9 ·
a
b
·
b
a
+ 10
= 2
9 + 10
16
Question 6.2.10. x > 0, y > 0 , 2x + 3y = 1 ,
1
x
+
1
y
.
6.2 33
Solution.
1
x
+
1
y
= (2x + 3y)(
1
x
+
1
y
)
= 2x ·
1
x
+ 2x ·
1
y
+ 3y ·
1
x
+ 3y ·
1
y
= 2 + 2 ·
x
y
+ 3 ·
y
x
+ 3
2
r
2 ·
x
y
· 3 ·
y
x
+ 5
= 2
6 + 5
2
6 + 5
Question 6.2.11. x > 0, y > 0, z > 0 , x + 2y + 3z = 1 ,
1
x
+
2
y
+
3
z
.
Solution.
1
x
+
2
y
+
3
z
= (x + 2y + 3z)(
1
x
+
2
y
+
3
z
)
= 1 + 2 ·
x
y
+ 3 ·
x
z
+ 2 ·
y
x
+ 4 + 6 ·
y
z
+ 3 ·
z
x
+ 6 ·
z
y
+ 9
= 14 + 2 ·
x
y
+ 3 ·
x
z
+ 2 ·
y
x
+ 6 ·
y
z
+ 3 ·
z
x
+ 6 ·
z
y
14 + 3
3
r
2 ·
x
y
· 6 ·
z
y
· 3 ·
x
z
+ 3
3
r
3 ·
z
x
· 2 ·
x
y
· 6 ·
y
z
= 14 + 3
3
36 + 3
3
36
14 + 6
3
36
Question 6.2.12. x > 0, y > 0 ,
2y
x
2
+
3x
y
+ 36x .
Solution.
2y
x
2
+
3x
y
+ 36x 3
3
r
2y
x
2
·
3x
y
· 36x
= 3
3
6 · 36
= 3 · 6 = 18
18
Question 6.2.13. x > 0 , x
2
+
2
x
.
Solution.
x
2
+
2
x
= x
2
+
1
x
+
1
x
3
3
r
x
2
·
1
x
·
1
x
= 3
3
Question 6.2.14. x > 0, y > 0 2x + 3y = 5 ,
2x +
3y .
6.2 34
Solution.
2x + 3y 2
p
2x · 3y = 2
p
6xy
5 2
p
6xy
2
p
6xy 5 (6.4)
(
2x +
p
3y)
2
= 2x + 2
p
6xy + 3y
= 2x + 3y + 2
p
6xy
= 5 + 2
p
6xy (6.5)
(6.4) (6.5)
(
2x +
p
3y)
2
= 5 + 2
p
6xy
5 + 5 = 10
2x +
p
3y
10
10
Question 6.2.15. x > 0 , x +
1
x
+
4x
x
2
+ 1
.
Solution.
x +
1
x
+
4x
x
2
+ 1
=
x
2
+ 1
x
+
4x
x
2
+ 1
2
s
x
2
+ 1
x
·
4x
x
2
+ 1
= 2
4 = 4
4.
Question 6.2.16. 2 이의 .
Solution. , 밑변 a, b, r
.
, a · b . ,
a
2
+ b
2
2
a
2
· b
2
(2r)
2
2ab
2r
2
ab
2 · 4 ab
8.
6.2 35
Question 6.2.17. 2 .
Solution. , , 각각 a, b, c .
r(= 2) . abc .
2 . 2r .
, a
2
+ b
2
+ c
2
= (2r)
2
= 2
2
r
2
. ,
2
2
r
2
= a
2
+ b
2
+ c
2
3
3
a
2
b
2
c
2
3, 2
6
r
6
3
3
(abc)
2
2
6
· 2
6
3
3
(abc)
2
abc
2
6
3
3
2
6
3
3
Question 6.2.18. x
2
+ y
2
= 4 , 3x + 4y 값과 . , ,
x, y 각각 .
Solution. By Cauchy-Shwarz inequality,
(3
2
+ 4
2
)(x
2
+ y
2
) (3x + 4y)
2
(6.6)
6.2 36
(9 + 16)(4) (3x + 4y)
2
100 = 25 · 4 (3x + 4y)
2
10 3x + 4y 10 (6.7)
: 10, : -10
x
3
=
y
4
,
(10) , (
6
5
,
8
5
)
(-10) , (
6
5
,
8
5
)
Question 6.2.19. x
2
+ y
2
+ z
2
= 1 , 2x 3y + 6z 값과 .
Solution. By Cauchy-Shwarz inequality,
(2
2
+ (3)
2
+ 6
2
)(x
2
+ y
2
+ z
2
) (2x 3y + 6z)
2
(4 + 9 + 36) ·1 (2x 3y + 6z)
2
49 (2x 3y + 6z)
2
7 2x 3y + 6z 7
: -7, : 7
Question 6.2.20. 3x
2
+ 2y
2
= 2 , 2x 5y 값과 .
Solution. By Cauchy-Shwarz inequality,
((
2
3
)
2
+ (
5
2
)
2
)((
3x)
2
+ (
2y)
2
) (2x 5y)
2
(
4
3
+
25
2
) · 2 (2x 5y)
2
8 + 75
6
· 2 (2x 5y)
2
83
3
(2x 5y)
2
83
3
2x 5y
83
3
:
83
3
, :
83
3
Question 6.2.21. 6 P . P
a, b, c , a
2
+ b
2
+ c
2
. , a, b, c 각각 .
6.2 37
Solution.
1
2
· 6 · 3
3 =
1
2
· 6 · a +
1
2
· 6 · b +
1
2
· 6 · c =
1
2
· 6 · (a + b + c) .
(a + b + c) = 3
3.
(a
2
+ b
2
+ c
2
)(1
2
+ 1
2
+ 1
2
) (a + b + c)
2
= (3
3)
2
(a
2
+ b
2
+ c
2
) 27 (1)
(1) ,
a
1
=
b
1
=
c
1
어야 , , a : b : c = 1 : 1 : 1
a = b = c =
3
7
7.1
7.1.1
Definition 7.1.1 (domain and image). Let R be a relation from A to B. The domain of the ralation
R, denoted by Dom(R), is the set of all those a A; and the image of R, denoted by Img(R), is the
set of all those b B such that aRb for some a A. In symbols,
Dom(R) = {a A | b B, (a, b) R}
and
Img(R) = {b B | a A, (a, b) R}
Definition 7.1.2 (function). Let X and Y be sets. A function from X to Y is a triple (f, X, Y ), where
f is a relation from X to Y satisfying
(a) Dom(f) = X.
(b) If (x, y) f and (x, z) f, then y = z.
Definition 7.1.3 (identical function). f g (Symbolically) f = g
Dom(f) = Dom(g)
Ran(f) = Ran(g)
x Dom(f), f(x) = g(x)
Example 7.1.1 ( ).
Dom = {1, 2}, f(x) = x, g(x) = x
3
Definition 7.1.4 (). Let f: X Y be a function. f
1
x
1
6= x
2
y
1
6= y
2
Definition 7.1.5 ().
2
Ran(f) = Img(f)
Definition 7.1.6 (= ).
3
Definition 7.1.7 (). f : X X f (x) = x
1
injective, one-to-one,
2
surjective, onto
3
bijective,
7.1 39
Definition 7.1.8 (). f f(x) = c
Definition 7.1.9 (). x
1
< x
2
f (x
1
) f (x
2
)
Definition 7.1.10 (). x
1
< x
2
f (x
1
) < f (x
2
)
Definition 7.1.11 ().
x X, f(x) = f(x)
Definition 7.1.12 ().
x X, f(x) = f(x)
Definition 7.1.13 (x = a ).
x X, f(2a x) = f(x)
x X, f(a x) = f(a + x)
Definition 7.1.14 (y = b ).
x X, 2b f(x) = f(x)
x X, (f(x) b) = f(x) b
Definition 7.1.15 ( (a, b) ).
x X, 2b f(2a x) = f (x)
x X, 2b = f(x) + f (2a x)
x X, 2b = f(a x) + f (a + x)
Definition 7.1.16 ().
안에 반복 , x f(x) = f(x + p) 최초 p
.
Note ( ).
Π(n, m)
n
P
m
S(m, n) ·
n
P
n
n
P
n
n
C
m
n
H
m
Theorem 7.1.1 ( ). .
X, Y, n(X) = m, n(Y) = n f : X Y
n
Π
m
= n
m
X, Y , n(X) = m, n(Y ) = n, m n f : X Y
n
P
m
=
n!
(n m)!
7.1 40
X, Y , n(X) = m, n(Y ) = n, m n f : X Y
S(m, n) · P er(n, n)
X, Y , n(X) = m, n(Y ) = n, m n f : X Y
=
( 1 ) × C(n, 1)
( 2 ) × C(n, 2)
( 3 ) × C(n, 3)
( 4 ) × C(n, 4)
( n-1 ) × C(n, n 1)
= Π(n, m)
Π(1, m) × C(n, 1)
(Π(2, m) C(2, 1)) × C(n, 2)
(Π(3, m) (C(3, 1) + (Π(2, r) C(2, 1)) × C(n, 2)) × C(n, 3)
···
( n-1 ) × C(n, n 1)
X, Y , n(X) = m, n(Y ) = n, m = n f : X Y
n
P
m
= n!
X, Y , n(X) = m, n(Y ) = n f : X Y
n
C
m
=
P (n, m)
m!
X, Y , n(X) = m, n(Y ) = n f : X Y
n
H
m
= C((n 1 + m), m)
Example 7.1.2 ( 1). X, Y , n(X) = 5, n(Y ) = 3 f : X Y
Solution.
S(5, 3) · P er(3, 3)
25 · 3!
150
7.1 41
Solution. another solution ; n = 3, m = 5 ,
Π(3, 5)
Π(1, 5) × C(3, 1)
(Π(2, 5) C(2, 1)) × C(3, 2)
= 3
5
1
5
× 5
(2
5
2) × 3
= 243
5
90
= 150
Example 7.1.3 ( 2). X, Y , n(X) = 4, n(Y ) = 3 f : X Y
Solution.
S(4, 3) · P er(3, 3)
= C(4, 2) · C(2, 2) · C(1, 1) ·
1
2
· 3!
= 6 · 2 ·
1
2
· 3!
= 36
Solution. n = 3, m = 4 ,
Π(3, 4)
Π(1, 4) × C(3, 1)
(Π(2, 4) C(2, 1)) × C(3, 2)
= 3
4
1
4
× 3
(2
4
2) × 3
= 81
3
42
= 36
Example 7.1.4 ( ). X, Y , n(X) = 3, n(Y ) = 4 f : X Y
7.1 42
Solution. n = 4, m = 3 ,
C(4, 3)
= 4
Example 7.1.5 ( ). X, Y , n(X) = 3, n(Y ) = 4 f : X Y
Solution. n = 4, m = 3 ,
H(4, 3)
= C(4 + 3 1, 3)
= C(6, 3)
= 20
Question 7.1.1. X = {1, 2, 3}, Y = {3, 4, 5, 6, 7} , f : X Y f ?
Solution.
1. ,
n
Π
r
,
5
Π
3
= 5
3
= 125
2. ,
X = {1, 2, 3} , 1 Y 5 , 2
Y 5 있으, 3 Y 5 .
5
3
= 125 .
Question 7.1.2. X = {1, 2, 3}, Y = {3, 4, 5, 6, 7} , f : X Y f
?
Solution.
1. ,
n
P
r
,
5
P
3
= 5 · 4 · 3 = 60
2. , X Y
응은 ( )어야 ,
X = {1, 2, 3} , , 1 Y 5
, 2 Y 4 있으, 3 Y 3
.
5
P
3
= 5 · 4 · 3 = 60 .
Question 7.1.3. X = {a, b, c, d, e}, Y = {1, 2, 3} , f : X Y
f ?
Solution.
3
Π
5
= 3
5
= 243 (7.1)
(7.1) Y Y 1 Y 2
어야 . ,
3
Π
5
(1 · 3 + (
2
Π
5
2) · 3)
243 (3 + (32 2) ·3)
7.1 43
243 (3 + 90) = 150 (7.2)
150 ().
Question 7.1.4. X = {1, 2, 3}, Y = {3, 4, 5, 6, 7} , f : X Y f
? (, x
1
< x
2
f (x
1
) < f (x
2
) )
Solution.
1. ,
n
C
r
,
5
C
3
=
5!
2!3!
= 10
2. ,
Y = {3, 4, 5, 6, 7} 5 X = {1, 2, 3} ,
3 어야 . 5 3
3 3 ·2 ·1 = 6
1 있으,
5 · 4 · 3
3 · 2 · 1
= 10 .
Question 7.1.5. X = {−1, 0, 1} , f : X Y 각각
.
Solution.
1. f(x) = f(x)
f(1) = f(1) = 1 (f (0) = 1 f(0) = 0 f (0) = 1) , 3
f(1) = f(1) = 0 (f (0) = 1 f(0) = 0 f (0) = 1) , 3
f(1) = f(1) = 1 (f (0) = 1 f(0) = 0 f (0) = 1) , 3
9
2. f(x) = f(x)
f(0) = f(0) f(0) = f (0) 2 · f (0) = 0
f (0) = 0
, , .
f(1) = 1 f(1) = 1 , 1
f(1) = 0 f(1) = 0 , 1
f(1) = 1 f(1) = 1 , 1
3
Question 7.1.6. f (x) x
1
, x
2
, x
3
f(
x
1
+ x
2
+ x
3
3
) <
f(x
1
) + f(x
2
) + f(x
3
)
3
임을 .
7.1 44
Solution. Jensen inequality .
. 방법 있을 .
Question 7.1.7. f(x) = 5(x 1)
2
(x 3)
2
+ 2 , x f(x) = k
k 각각 .
Solution.
...
Question 7.1.8. f : R R 실수 . f(x) + f(x) = 0 f(3) =
2, f(1) = 5 , f(0) + f(1) + f(3) .
Solution. f(x) + f(x) = 0 f(x) = f(x). ,
f(0) + f(1) + f(3)
= f(0) f(1) f(3)
= 0 (5) 2
= 3
3
Question 7.1.9. f : R R 실수 . f(x) + f(4 x) = 6
f(0) = 2 , f(2) + f(4) .
Solution.
1. , x .
2. ,
f(x) + f(4 x) = 6
f(4 x) = 6 f(x)
f(2 · 2 x) = 2 · 3 f (x)
, f(x) (2,3) . , f(2) = 3
,
f(4 0) = 6 f(0)
f(4) = 6 f(0)
= 6 (2)
= 8
f (2) + f (4) = 3 + 8 = 11
Question 7.1.10. f(x) = x
2
(1 x 1) f(x) = f(x + 2) , f(1) + f (2) + ··· + f(100)
.
7.1 45
Solution.
f(1) + f(2) + ··· + f (100)
= f(1) + f(1) + f(1) + f(1) + ··· + f(1) + f(1)
= 50 · (f(1) + f(1))
= 50 · (1 + 1) = 100
100
Question 7.1.11. y = f(x) f(x) = f(2 x), f (4 x) = f (4 + x) , y = f(x)
p . , p .
Solution.
f(4 x) = f(4 + x)
f(4 (2 x)) = f(4 + x)
f(x + 2) = f(x + 4)
x
0
= x + 2 f(x
0
) = f(x
0
+ 2)
p = 2
Question 7.1.12. X = {1, 2, 3, 4, 5, 6}, Y = {4, 5, 6, 7, 8, 9, 10, 11}, f : X Y . f(3) = 7
, x
1
< x
2
f (x
1
) < f (x
2
) f .
Solution. f(3) = 7 , {1, 2} {4, 5, 6} {4, 5, 6, 7}
{8, 9, 10, 11} . ,
1. {1, 2} {4, 5, 6}
3
C
2
=
3
C
1
= 3
2. {4, 5, 6, 7} {8, 9, 10, 11}
4
C
4
=
4
C
0
= 1
4.
Question 7.1.13. f(x) = f (x + 2), g(x) = g(3 x) , k(x) = f(x) + g(x) p
. p .
Solution.
g(x + 2) = g(3 (x + 2)) = g(1 x) (7.3)
,
7.1.2
Theorem 7.1.2 ().
1. h(x) = f
1
(x) f (f
1
(x)) = f
1
(f(x)) f (h(x)) = h(f (x)) = x
2. h(x) = (f g)
1
(x) (f g)((f g)
1
(x)) = (f g)
1
((f g)(x)) (f g)(h(x)) = h((f g)(x)) =
x
3. h(x) = (f (x + a))
1
(x) (f (x + a))(h(x)) = h(f (x + a)) = x
7.2 46
Note (알아 ).
f y = x f f
1
( 는다.)
Note (알아 ).
(f : X X) F f f = f
1
1. f f f = I
X
2. f(a) = b f(b) 6= a
Proof.
1. f f f = f
1
f = I
X
2.
1) f(b) = a
2) f
1
(a) = b f(f(a)) = b f(b) = b
N
Note ( ).
1. (f f f = I) (f(a) = b f (b) 6= a)
2. (f f = I) (f(a) = b f (b) = a)
3. (f f = f) (f(a) = b f(b) = b)
7.2
7.2.1
Definition 7.2.1 (). A, B(B 6= 0)
A
B
. B 0
A
B
.
Theorem 7.2.1 ( , proportional expression). a : b = c : d (
a
b
=
c
d
) ,
1. ad = bc
2.
a + b
b
=
c + d
d
3.
a b
b
=
c d
d
4.
a + b
a b
=
c + d
c d
5. ( )
a
b
=
c
d
=
e
f
a
b
=
c
d
=
e
f
=
a + c + e
b + d + f
=
pa + qc + re
pb + qd + rf
7.2 47
Proof.
a
b
=
c
d
=
e
f
a
b
=
c
d
=
e
f
=
pa
pb
=
qc
qd
=
re
rf
=
pa + qc + re
pb + qd + rf
7.2.2
8
8.1
8.1.1 곱과
Definition 8.1.1 (an nth root of a number x). an nth root of a number x, where n is usually assumed
to be a positive integer, is a number r which, when raised to the power n yields x:
r
n
= x,
where n is the degree of the root. A root of degree 2 is called a square root and a root of degree 3, a
cube root. Roots of higher degree are referred by using ordinal numbers, as in fourth root, twentieth
root, etc.
8.1.2
8.2
Theorem 8.2.1 (Misc).
a
logax
= b
logbx
x =
1
ab
a
logb
= b
loga
8.3
Definition 8.3.1 ( ).
logN = z + α (where z Z 0 α < 1)
z : 부분
α : 부분
log20 = 1 + 0.3010 = 1.3010
log0.0002 = 4 + 0.3010 =
¯
4.3010
2.699 6=
¯
2.699
2.699 = 2 0.699 6= 2 + 0.699 =
¯
2.699
Note ( 부분()).
음을 부분 릿 해하 .
8.4 49
1. log10 = 1 : 2, 부분 1
2. log10
1
=
¯
1 : 1, 부분 -1
Note ( 부분()).
음을 부분 자의 해하 .
N logN N = a ·10
z
(where 1 a < 10 z Z) .
, N
0
(z
0
Z) N
0
= a · 10
z
0
N N
0
부분 .
8.4
8.4.1
Definition 8.4.1 (). x R , a
x
,
y = a
x
(a > 0, a 6= 1)
x . a .
Question 8.4.1.
lim
x→∞
x
n
e
x
=?
Solution. We know that e
x
=
P
k=0
x
k
k!
x
n
e
x
=
x
n
e
x
=
x
n
P
n
k=0
x
k
k!
+
P
k=n+1
x
k
k!
. We conclude that lim
x→∞
x
n
e
x
= 0
8.4.2
Definition 8.4.2 (). y = a
x
(a > 0, a 6= 1)
y = log
a
x (a > 0, a 6= 1)
. a .
8.4.3
Theorem 8.4.1 ( ). y = a
f(x)
.
y
0
= a
f(x)
· f
0
(x)
Proof. Let u = f(x), then
dy
dx
=
dy
du
·
du
dx
.
dy
du
= a
u
· ln a
So
dy
dx
=
d
dx
a
f(x)
= a
f(x)
· ln a · f
0
(x)
8.4 50
Theorem 8.4.2 ( ). y = log f(x) .
d
dx
log
a
f(x) =
1
f(x) · ln a
· f
0
(x)
Proof. Let u = f(x), then
dy
dx
=
dy
du
·
du
dx
.
dy
du
=
1
u · ln a
So
dy
dx
=
d
dx
log
a
f(x) =
1
f(x) · ln a
· f
0
(x)
9
9.1
9.1.1
Definition 9.1.1 ( (radius vector)). XOP OX
O OP XOP .
OX , OP .
Figure 9.1:
Definition 9.1.2 ( ). OP O
, 음의 ,
(+), 음의 음의 (-) .
Figure 9.2:
Note ( ).
9.1 52
Figure 9.3:
9.1.2
Definition 9.1.3 (). 360 1, 1 1/60
1, 1 1/60 1 방법
Definition 9.1.4 (1 radian). 1 radian ,
57
17
0
45
00
.
9.1.3
Definition 9.1.5 ( ).
1. : 한하
2. :
Note ( ).
n
1. sinx = a(|a| 1) α ( i.e. sinx = sinα )
x = (1)
n
α +
2. cosx = a(|a| 1) α ( i.e. cosx = cosα )
x = ±α + 2
9.2 53
3. tanx = a α
x = α +
9.2
9.2.1
Theorem 9.2.1 ( ).
(f(x) = asinx + ocosx) (Max(f ) = f (α) tanα =
a
o
)
Question 9.2.1.
f(x) = 3sinx + 4cosx x = α , tanα . (, 0 x π)
Solution.
3
4
9.3
Note ( ).
1.
2.
Note ( 2 ).
1.
2.
3.
9.3.1
Definition 9.3.1 ( 6). ABC A, B, C A, B, C ,
BC, CA, AB 각각 a, b, c . ,
A, B, C, a, b, c 6 .
Theorem 9.3.1 ( ).
: ABC R ABC
계가 .
a
sinA
=
b
sinB
=
c
sinC
= 2R
sinA =
a
2R
, sinB =
b
2R
, sinC =
c
2R
9.3 54
a = 2RsinA, b = 2RsinB, c = 2RsinC
a : b : c = sinA : sinB : sinC
1
a = bcosC + ccosB
b = ccosA + acosC
c = acosB + bcosA
2
a
2
= b
2
+ c
2
2bccosA
b
2
= c
2
+ a
2
2cacosB
c
2
= a
2
+ b
2
2abcosC
cosA =
b
2
+ c
2
a
2
2bc
cosB =
c
2
+ a
2
b
2
2bc
cosC =
a
2
+ b
2
c
2
2bc
Theorem 9.3.2 ( ).
1. S =
1
2
absinC =
1
2
bcsinA =
1
2
casinB
2. R ,
S =
abc
4R
= 2R
2
sinAsinBsinC
3. r ,
S =
1
2
r(a + b + c)
4. : s =
a + b + c
2
,
S =
p
s(s a)(s b)(s c)
Theorem 9.3.3 ( ).
1.
sin
2
θ
2
=
1 cosθ
2
cos
2
θ
2
=
1 + cosθ
2
tan
2
θ
2
=
1 cosθ
1 + cosθ
2. tan
θ
2
= t,
9.3 55
sinθ =
2t
1 + t
2
cosθ =
1 t
2
1 + t
2
tanθ =
2t
1 t
2
Theorem 9.3.4 (3).
sin3α = 3sinα 4sin
3
α
cos3α = 4cos
3
α 3cosα
Theorem 9.3.5 ( ).
sinαcosβ =
1
2
{sin(α + β) + sin(α β)}
cosαsinβ =
1
2
{sin(α + β) sin(α β)}
cosαcosβ =
1
2
{cos(α + β) + cos(α β)}
sinαsinβ =
1
2
{cos(α + β) cos(α β)}
Theorem 9.3.6 ( ).
고고
S+S=2SC
S-S=2CS
C+C=2CC
C-C=-2SS
sinA + sinB = 2sin
A + B
2
cos
A B
2
sinA sinB = 2cos
A + B
2
sin
A B
2
cosA + cosB = 2cos
A + B
2
cos
A B
2
cosA cosB = 2sin
A + B
2
sin
A B
2
9.4 56
9.3.2
9.3.3
9.4
Theorem 9.4.1 ( ). .
1. sin(α + β) = sinαcosβ + cosαsinβ
sin(α β) = sinαcosβ cosαsinβ
2. cos(α + β) = cosαcosβ sinαsinβ
cos(α β) = cosαcosβ + sinαsinβ
3. tan(α + β) =
tanα + tanβ
1 tanαtanβ
tan(α β) =
tanα tanβ
1 + tanαtanβ
Theorem 9.4.2 ( ). c sin θ + d cos θ(c 6= 0, d 6= 0)
.
a sin θ + o cos θ =
p
a
2
+ o
2
sin(θ + β) (where sin β =
o
a
2
+ o
2
, cos β =
a
a
2
+ o
2
) (9.1)
o sin θ + a cos θ =
p
a
2
+ o
2
cos(θ β) (where sin β =
o
a
2
+ o
2
, cos β =
a
a
2
+ o
2
) (9.2)
Theorem 9.4.3 ( ).
sin2x = 2sinxcosx =
2tanx
1 + tan
2
x
cos2x = cos
2
x sin
2
x = 2cos
2
x 1 = 1 2sin
2
x =
1 tan
2
x
1 + tan
2
x
tan2x =
2tanx
1 tan
2
x
Theorem 9.4.4 (3 ).
sin3x = 3sinx 4sin
3
x
cos3x = 4cos
3
x 3cosx
tan3x =
3tanx tan
3
x
1 3tan
2
x
Theorem 9.4.5 ( ).
cos
2
x
2
=
1 + cosx
2
sin
2
x
2
=
1 cosx
2
tan
2
x
2
=
1 cosx
1 + cosx
10
10.1
Definition 10.1.1 ( ). 례로 ,
.
a
1
, a
2
, a
3
, ..., a
n
, ...
, 앞에 례로 , , , ..., n, ... 1, 2, 3, ...,
n, ... .
10.2
Definition 10.2.1 (). 례로
, .
a
1
, a
2
, a
3
, ..., a
n
, a
n+1
, ... a, d
. a
1
= a
a
n+1
a
n
= d a
n+1
= a
n
+ d (, n = 1, 2, 3, ...)
Definition 10.2.2 ( ). a, d ( n) a
n
a
n
= a + (n 1)d
Definition 10.2.3 (). a, b, c b a, c
.
a, b, c 2b = a + c b = (a + c)/2
a, c a, c
Definition 10.2.4 ().
. ,
1
a
1
,
1
a
2
,
1
a
3
, ...,
1
a
n
a
1
, a
2
, a
3
, ..., a
n
Definition 10.2.5 (). 0 a, b, c b a, c
.
Theorem 10.2.1. b a, c ,
2
b
=
1
a
+
1
c
x =
2ac
a + c
Theorem 10.2.2. b a, c ,
10.2 58
ac < b
2
1
a
+
1
b
>
2
b
10.2.1
Let a > 0, b > 0.
() - an arithmetical mean
A =
a + b
2
() - a mean proportional (a geometric mean)
G =
ab
- a harmonic mean
H =
2ab
a + b
A G H
G
2
= AH
10.2.2
Theorem 10.2.3.
n S
n
1. a, n l ,
S
n
=
n(a + l)
2
2. a, d ,
S
n
=
{n(2a + (n 1)d}
2
10.2.3 이의
Theorem 10.2.4.
{a
n
} n S
n
음이 .
a
1
= S
1
, a
n
= S
n
S
n1
(n 2)
1. a, n l ,
S
n
=
n(a + l)
2
10.3 59
2. a, d ,
S
n
=
{n(2a + (n 1)d}
2
Theorem 10.2.5.
{a
n
} n S
n
S
n
= An
2
+ Bn + C (where A, B, C : Constant)
1. C = 0 {a
n
} .
2. C 6= 0 {a
n
} 2 .
10.3
10.3.1
Theorem 10.3.1.
a, r n S
n
1. r 6= 1 ,
S
n
=
a(1 r
n
)
1 r
=
a(r
n
11)
r 1
2. r = 1 ,
S
n
= na
Theorem 10.3.2.
{a
n
} n S
n
S
n
= Ar
n
+ B (where r 6= 0, r 6= 1, A, B : Constant)
1. A + B = 0 {a
n
} .
2. A + B 6= 0 {a
n
} 2 .
10.3.2
Definition 10.3.1 (). 이자 합한 .
Definition 10.3.2 (). , .
Definition 10.3.3 (期末). ,
.
Theorem 10.3.3 ( ). .
If r 6= 1, then S
n
=
a(1 r
n
)
1 r
If r = 1, then S
n
= na
10.3 60
Proof. To prove it by contradiction try and assume that the statement is false, proceed from there
and at some point you will arrive to a contradiction.
Theorem 10.3.4 (). 음의 a, ( ) 율을 r n
( n)
a(1 + r)
n
Theorem 10.3.5 ( -,). a, 이율을 r,
n, S
S = a(1 + rn)
S = a(1 + r)
n
Corollary 10.3.5.1 ( -). , 이율은 r a
n
S =
a(1 + r){(1 + r)
n
1}
r
Proof. S ,
S = a(1 + r) + a(1 + r)
2
+ a(1 + r)
3
+ ... + a(1 + r)
n
S =
a(1 + r){(1 + r)
n
1}
(1 + r) 1
=
a(1 + r){(1 + r)
n
1}
r
Corollary 10.3.5.2 ( -). , 이율은 r a
n
S =
a{(1 + r)
n
1}
r
Definition 10.3.4 (). , 액씩
, . , 액씩
, .
Theorem 10.3.6 (). A , 액씩 n
. ? , 이율 r, 1 .
Proof. A n A(1 + r)
n
···
, x n x + x(1 + r) + x(1 + r)
2
+
··· + x(1 + r)
n1
=
x{(1 + r)
n
1}
(1 + r) 1
···
아야
x{(1 + r)
n
1}
r
= A(1 + r)
n
x =
rA(1 + r)
n
(1 + r)
n
1
10.4 61
Definition 10.3.5 ( ).
. .
재의 .
Theorem 10.3.7 ( ). a n .
는다 마를 아야 ? , 이율 r, 1 .
Proof. P ,
P (1 + r)
n
= a + a(1 + r) + a(1 + r)
2
+ ··· + a(1 + r)
n1
P (1 + r)
n
=
a{(1 + r)
n
1)}
(1 + r) 1
P =
a{(1 + r)
n
1)}
r(1 + r)
n
Theorem 10.3.8 (). n fv(future value) 는다. 1 n
fv 는다. 이율 r, 1 . fv (pv)?
Proof. 1 n .
1 : fv = pv + pv ×r = pv × (1 + r)
n : fv = pv × (1 + r)
n
pv =
fv
(1 + r)
n
10.4
Definition 10.4.1 (
P
).
n
X
k=1
a
k
= a
1
+ a
2
+ ··· + a
n1
+ a
n
Theorem 10.4.1 ( ).
P
n
k=1
k =
n(n + 1)
2
P
n
k=1
k
2
=
n(n + 1)(2n + 1)
6
P
n
k=1
k = {
n(n + 1)
2
}
2
Theorem 10.4.2 ( ).
P
n
k=1
k(k + 1) =
n(n + 1)(n + 2)
3
10.4 62
P
n
k=1
k(k + 1)(k + 2) =
n(n + 1)(n + 2)(n + 3)
4
P
n
k=1
k(k + 1)(k + 2)(k + 3) =
n(n + 1)(n + 2)(n + 3)(n + 4)
5
Proof.
Figure 10.1:
Theorem 10.4.3 ( ).
10.5 63
n
X
k=1
1
k +
k + 1
=
n
X
k=1
k + 1
k
Theorem 10.4.4 (() · () ).
1 ·5 + 2 ·5
2
+ 3 ·5
3
+ ···+ n ·5
n
+ ···
.
1. S 는다.
2. r , S rS . (, r 6= 1)
3. S .
Definition 10.4.2 ().
.
Theorem 10.4.5 ().
.
Figure 10.2:
열에 .
1. 는다.
2. n .
3. n .
4. k .
Theorem 10.4.6 ( a
n
= f(n + 1) f(n) ).
S
n
=
n
X
k=1
(f(k + 1) f(k)) = f(n + 1) f (1)
10.5
Definition 10.5.1 ().
() (, Recurrence relation) 이의
. , {a
n
} a
n
f
a
n+1
= f(a
n
)
있을 , f {a
n
} , , {a
n
}
f .
10.5 64
{a
n
} a
n
n (explicit formula)
.
Definition 10.5.2 ( ). {a
n
}
1. a
1
2. a
n
, a
n+1
이의 where n N
이의
.
10.5.1
10.5.2
10.5.3
Theorem 10.5.1 ( ).
1. a
n+1
= a
n
+ f(n)
a
n
= a
1
+ f(1) + f(1) + f(2) + ··· + f(n 1) = a
1
+
n1
X
k=1
f(k)
2. a
n+1
= a
n
· f(n)
a
n
= a
1
· f(1) · f(1) · f(2) · ··· · f(n 1) = a
1
·
n1
Y
k=1
f(k)
11
11.1
Definition 11.1.1 ( ). {a
n
} a
n
, a
n
α
{a
n
} α . α {a
n
} ,
lim
n→∞
a
n
= α n a
n
α
.
Definition 11.1.2 ( ). {a
n
} , .
Note. {a
n
} 3 .
1.
lim
n→∞
a
n
= n a
n
2. 음의
lim
n→∞
a
n
= −∞ n a
n
−∞
3. {a
n
} 음의 {a
n
}
.
11.2
Theorem 11.2.1 ( ). {a
n
}, {a
n
} lim
n→∞
a
n
= α,
lim
n→∞
b
n
= β (α, β 실수) , 음이 .
lim
n→∞
(a
n
+ b
n
) = lim
n→∞
a
n
+ lim
n→∞
b
n
lim
n→∞
(a
n
b
n
) = lim
n→∞
a
n
lim
n→∞
b
n
lim
n→∞
(c · a
n
) = c · lim
n→∞
a
n
(when c is constant)
lim
n→∞
(a
n
· b
n
) = lim
n→∞
a
n
· lim
n→∞
b
n
lim
n→∞
(a
n
/b
n
) = lim
n→∞
a
n
/ lim
n→∞
b
n
(when b
n
, lim
n→∞
b
n
6= 0)
11.3 66
11.3
11.4
Theorem 11.4.1 ( ). {a
n
}, {a
n
} lim
n→∞
a
n
= α, lim
n→∞
b
n
= β
(α, β 실수) , 음이 .
(n N)(a
n
b
n
α β)
( {c} , n N)(a
n
c
n
b
n
α = β lim
n→∞
c
n
= α)
11.5
Theorem 11.5.1 ( ). 1, r, a
n
= r
n1
{a
n
} r .
r > 1 lim
n→∞
r
n
=
r = 1 lim
n→∞
r
n
= 1
|r| < 1 lim
n→∞
r
n
= 0
r 1 lim
n→∞
r
n
i.e.
Theorem 11.5.2 ( ). a, r, a
n
= ar
n1
{a
n
} .
(a = 0) (1 < r 1)
11.6
11.7 이의
11.8
11.9
12
12.1
12.1.1
12.1.2
12.1.3
12.1.4
Definition 12.1.1 (the limit of a function). f : D R be a function defined on a subset D R, let
a be a limit point of D, and let L be a real number. Then the function f has a limit L at a
is defined to mean
for all ε > 0, there exists a δ > 0 such that for all x in D that satisfy 0 < |x c| < δ, the inequality
|f(x) L| < ε holds.
Symbolically:
lim
xa
f(x) = L (ε > 0)( δ > 0)(x D)(0 < |x a| < δ |f(x) L| < ε)
Definition 12.1.2 (One-sided limit, right-sided limit).
lim
xa
+
f(x) (ε > 0)( δ > 0)(x D)(0 < x a < δ |f(x) L| < ε)
Definition 12.1.3 (One-sided limit, left-sided limit).
lim
xa
f(x) (ε > 0)( δ > 0)(x D)(0 < a x < δ |f(x) L| < ε)
Theorem 12.1.1 ( ). lim
xp
f(x) lim
xp
g(x) ,
lim
xp
(f(x) + g(x)) = lim
xp
f(x) + lim
xp
g(x)
lim
xp
(f(x) g(x)) = lim
xp
f(x) lim
xp
g(x)
lim
xp
(f(x) · g(x)) = lim
xp
f(x) · lim
xp
g(x)
lim
xp
(f(x)/g(x)) = lim
xp
f(x)/ lim
xp
g(x), when lim
xp
g(x) 6= 0
Theorem 12.1.2 ( ). lim
xa
f(x)
g(x)
= L (L ) ,
lim
xa
f(x) = 0 L 6= 0 lim
xa
g(x) = 0 (12.1)
lim
xa
g(x) = 0 lim
xa
f(x) = 0 (12.2)
12.2 68
12.2
12.2.1
Definition 12.2.1 (Continuity). A function f is said to be continuous at c if it is both defined at c
and its value at c equals the limit of f as x approaches c.
lim
xc
f(x) = f(c)
12.2.2
Theorem 12.2.1 ( ). f (x), g(x) x = c
x = c .
1. kf(x), k
2. f(x) ± g(x)
3. f(x)g(x)
4.
f(x)
g(x)
, g(x) 6= 0
Theorem 12.2.2 (EVT : The Maximum Minimum Theorem).
f C
0
([a, b], R) (x [a, b])(c [a, b])(d [a, b])(f(c) f (x) f(d) f(x))
Theorem 12.2.3 (EVT : · ).
f(x) [a, b] f (x) 값과 는다.
12.2.3
Theorem 12.2.4 (IVT : The intermediate value theorem).
f C
0
([a, b], R) f(a) 6= f(b) (k (Min(f(a), f(b)), Max(f(a), f(b)))(c (a, b))(f(c) = k)
Theorem 12.2.5 ( ).
f(x) [a, b] , f(a) 6= f (b) f(a) f(b) 이의 임의의 실수 k
f(c) = k c (a, b) .
Corollary 12.2.5.1 (Bolzano’s theorem).
(f C
0
([a, b], R) (f(a)f(b) < 0)) (c (a, b))(f(c) = 0)
Corollary 12.2.5.2 ( ). f(x) [a, b] , f (a)f (b) < 0
f(x) = 0 (a, b) 는다.
13
13.1
13.1.1
Definition 13.1.1 ().
y = f(x) x a b , y f(a) f(b) . x
b a x , y f(b) f(a) y , 각각 x, y
. ,
x = b a
y = f (b) f (a) = f (a + x) f(a)
Definition 13.1.2 ((ARC or MV) : Average Rate of Change).
ARC(f, a, b) = MV (f, a, b) =
y
x
=
f(b) f(a)
b a
=
f(a + x) f(a)
x
IRC : Instantaneous Rate of Change
Definition 13.1.3 (x=a , , IRC).
IRC(f, a) = f
0
(a) = lim
x0
f(a + x) f(a)
x
= lim
xa
f(x) f(a)
x a
IRC : Instantaneous Rate of Change
IRC(f, a) = f
0
(a) = lim
x0
ARC(f, a, a + x)
Definition 13.1.4 ( ).
y = f(x) x y = f(x)
.
y = f (x) 역에 x y = f(x)
.
Note ( ).
f(x) x = α
1. f(x)(x α) x = α , x = α 분불
2. f(x)(x α)(x α) x = α
13.1 70
f(x) x = α 분불
1. f(x) x = α
2. f(x)(x α)(x α) x = α
13.1.2 하학
Note ().
f
0
(a) y = f (x) 위의 P (a, f(a)) .
13.1.3 속성
13.1.4
Definition 13.1.5 (-a derived function).
y = f (x) x f
0
(x)
f
0
: X R, f
0
(x) = lim
x0
f(x + x) f(x)
x
는다. f
0
(x) f(x) ,
f
0
(x), y
0
,
dy
dx
,
d
dx
f(x)
.
Definition 13.1.6 ().
y = f(x) f
0
(x) y = f(x) x ,
.
Theorem 13.1.1 ( y = x
n
(n ) ).
1. y = x
n
( n N) y
0
= nx
n1
2. y = c (c ) y
0
= 0
Theorem 13.1.2 ( 실수, , , ). f(x), g(x), h(x)
1. {kf(x)}
0
= kf
0
(x) (, k )
2. {f(x) + g(x)}
0
= f
0
(x) + g
0
(x)
3. {f(x) g(x)}
0
= f
0
(x) g
0
(x)
4. {f(x)g(x)}
0
= f
0
(x)g(x) + f(x)g
0
(x)
5. {f(x)g(x)h(x)}
0
= f
0
(x)g(x)h(x) + f(x)g
0
(x)h(x) + f(x)g(x)h
0
(x)
6. [{f(x)}
n
]
0
= n{f(x)}
n1
f
0
(x) (, n )
13.1 71
Theorem 13.1.3 ( ).
1. f(x)
Mod(f(x), (x a)
2
) = f
0
(a)(x a) + f(a) (13.1)
2. f(x)
Mod(f(x), (x a)
2
) = 0 f(a) = 0, f
0
(a) = 0 (13.2)
Proof.
(13.1)
f(x) (x a)
2
Q(x), mx + n
f(x) = (x a)
2
Q(x) + mx + n (1)
(1) x
f
0
(x) = 2(x a)Q(x) + (x a)
2
Q
0
(x) + m (2)
(1), (2) x = a
f(a) = ma + n, n = f(a) ma (3)
f
0
(a) = m (4)
(3), (4) mx + n = f
0
(a)x + f(a) af
0
(a) = f
0
(a)(x a) + f(a)
(13.2)
mx + n = 0, m = 0, n = 0 어야 (3), (4) f(a) = 0, f
0
(a) = 0
Theorem 13.1.4 ().
(even function)
0
= (odd function) : .
Figure 13.1: differentiation
Note ( ).
13.1 72
Theorem 13.1.5 ( - hakchin notation 1).
f(x) f
1
(x) = g(x) , x = f
1
(y) = g(y)
.
g
0
(y) · f
0
(x) = 1 (13.3)
g
0
(x) · f
0
(y) = 1 (13.4)
Proof.
g(f(x)) = x (g(f(x)))
0
= (x)
0
g
0
(f(x)) · f
0
(x) = 1 (where f
0
(x)) 6= 0)
Theorem 13.1.6 ( - hakchin notation2).
f(x) f
1
(x) = g(x) , y = f
1
(x)
.
f
0
(f
1
(x)) · (f
1
)
0
(x) = 1 (where f
0
(f
1
(x)) 6= 0) (13.5)
f
0
(y) · (f
1
)
0
(x) = 1 (where f
0
(f
1
(x)) 6= 0) (13.6)
g
0
(y) · f
0
(x) = 1 (13.7)
g
0
(x) · f
0
(y) = 1 (13.8)
Proof.
f(f
1
(x)) = x (f(f
1
(x)))
0
= (x)
0
f
0
(f
1
(x)) · (f
1
)
0
(x) = 1 (where f
0
(f
1
(x)) 6= 0)
Theorem 13.1.7 ( ).
f(x) f
1
(x) , y = f
1
(x) .
dy
dx
=
1
dx
dy
or (13.9)
(f
1
)
0
(x) =
1
f
0
(y)
(where f
0
(y) 6= 0) (13.10)
13.2 73
Proof.
f(f
1
(x)) = x f
0
(f
1
(x))(f
1
)
0
(x)) = 1
(f
1
)
0
(x) =
1
f
0
(f
1
(x))
=
1
f
0
(y)
y = f
1
(x) x = f(y)
dy
dx
= (f
1
)
0
(x),
dx
dy
= f
0
(y)
dy
dx
=
1
f
0
(y)
=
1
dy
dx
(where f
0
(y) 6= 0)
Proof. y = f
1
(x) x = f (y)
dx
dy
= f
0
(y). x = f(y) x
d
dx
(x) =
d
dx
(f(y)), 1 =
d
dy
(f(y)) ·
dy
dx
= f
0
(y) ·
dy
dx
dy
dx
=
1
f
0
(y)
=
1
dy
dx
(where f
0
(y) 6= 0)
Example 13.1.1 ( ). f(x) g(x), f(3x 5)
h(x) , h
0
(x) g
0
(x) ?
Solution. g(x) = f
1
(x) f (g(x)) = x, h(x) = (f(3x5))
1
f (3x5)(h(x)) =
x f (3h(x) 5) = x g(x) = 3h(x) 5 g
0
(x) = 3h
0
(x) h
0
(x) =
1
3
g
0
(x)
13.2
13.2.1
Theorem 13.2.1 (Rolle’s theorem).
f C
0
([a, b]) f C
1
((a, b)) f(a) = f(b) c (a, b)(f
0
(c) = 0)
Theorem 13.2.2 (Rolle’s theorem). If f : [a, b] R is a continuous function, differentiable on the
open interval (a, b), and f(a) = f(b), then there exists at least one c in the open interval (a, b) such
that
f
0
(c) = 0
13.2 74
This version of Rolle’s theorem is used to prove the mean value theorem, of which Rolle’s theorem is
indeed a special case. It is also the basis for the proof of Taylor’s theorem.
Theorem 13.2.3 (MVT).
f C
0
([a, b], R) f C
1
((a, b), R) c (a, b)(f
0
(c) =
f(b) f(a)
b a
)
Theorem 13.2.4 (MVT- ).
f C
0
([a, b]) f C
1
((a, b)) c (a, b)(f
0
(c) = ARC(f, a, b) = MV (f, a, b))
Theorem 13.2.5 ( (MVT) : the mean value theorem). Let f : [a, b] R be a continuous
function, and differentiable on the open interval (a, b), where a < b. Then there exists some c in (a, b)
such that
f
0
(c) =
f(b) f(a)
b a
Example 13.2.1 ( ).
f(x) f
0
(
a + b
2
) =
f(b) f(a)
b a
13.2.2
Note (notation idea for increasing). 음을 . Shorthand notation for ”increases” and ”de-
creases”
Definition 13.2.1 (). f
¨
I(R) x
1
< x
2
f (x
1
) < f (x
2
)
Definition 13.2.2 (). f
¨
I(R) x
1
< x
2
f (x
1
) > f (x
2
)
Definition 13.2.3 (). h > 0, f(x h) < f(a) < f(x + h) f(x) x = a
.
Definition 13.2.4 (). h > 0, f(x h) > f(a) > f(x + h) f(x) x = a
.
Theorem 13.2.6 ( 1). f C
1
((a, b)) (x (a, b))(f
0
(x) > 0) f
¨
I((a, b))
Theorem 13.2.7 ( 1). f C
1
((a, b)) (x (a, b))(f
0
(x) < 0) f
¨
I((a, b))
Theorem 13.2.8 ( 2). f C
1
((a, b)) f (x (a, b))(f
0
(x) 0) f
¨
I((a, b))
Theorem 13.2.9 ( 2). f C
1
((a, b)) f (x (a, b))(f
0
(x) 0) f
¨
I((a, b))
Theorem 13.2.10 ( 3). f C
1
((a, b)) f
¨
I((a, b)) f
0
(x) 0
Theorem 13.2.11 ( 3). f C
1
(I) f
¨
I(I) f
0
(x) 0
13.2.3
Definition 13.2.5 (, critical point).
a Dom(f) a is a critical point f
0
(a) = 0 f
0
(a) does not exist.
13.2 75
Note (Extrema & Critical point).
는다
O
X
계값 는다.
Definition 13.2.6 (, local Maximum). f(x) x = a x = a f(x)
, f(x) x = a , f(a) .
Definition 13.2.7 (, local Maximum). lim
xa
f(x) = f(a) h > 0, x (a h, a + h), f (a)
f(x) f(x) x = a , f(a) .
Definition 13.2.8 ( , strict local Maximum). lim
xa
f(x) = f(a) h > 0, x (a
h, a + h) {a}, f(a) > f(x) f(x) x = a , f(a)
.
Definition 13.2.9 ( 2, strict local Maximum2). lim
xa
f(x) = f(a) h R {0}, f(a +
h) < f (a) f (x) x = a , f(a) .
Definition 13.2.10 (, local minimum). f(x) x = a x = a f(x)
, f(x) x = a , f(a) .
Definition 13.2.11 (, local minimum). lim
xa
f(x) = f(a) h > 0, x (ah, a + h), f (a)
f(x) f(x) x = a , f(a) .
Definition 13.2.12 ( , strict local minimum). lim
xa
f(x) = f (a) h > 0, x (a
h, a + h) {a}, f(a) < f(x) f(x) x = a , f(a)
.
Definition 13.2.13 ( 2, strict local Maximum2). lim
xa
f(x) = f(a)h R{0}, f(a+
h) < f (a) f (x) x = a , f(a) .
Theorem 13.2.12 (값과 ). f(x) x = a a 함하
f
0
(a) = 0.
Note. 위의 는다.
Theorem 13.2.13 ( ). f (x) f
0
(a) = 0, x = a f
0
(x)
양에 음으 f(x) x = a , f(a). (
f
0
(a) = 0 니다. , )
Theorem 13.2.14 ( ). f (x) f
0
(a) = 0, x = a f
0
(x)
f(x) x = a , f(a). (
f
0
(a) = 0 니다. , )
Theorem 13.2.15 (f(x) ).
f(x) 는다. f
0
(x) 2 f (x) = 0 2 는다.
Note ().
.
Note ( ).
13.3 76
1 :
2
13.2.4 ,
13.3
- http://rigvedawiki.net/w/
-
- ,
Note (Reference Sites).
Theorem 13.3.1 ( ).
1. y = sin x y
0
= cos x
2. y = cos x y
0
= sin x
3. y = tan x y
0
= sec
2
x
4. y = csc x y
0
= csc x cot x
5. y = sec x y
0
= sec x tan x
6. y = cot x y
0
= csc
2
x
Note. 6 .
.
(Hyperbolic function)
13.4 77
Theorem 13.3.2 ( ).
y = f (x)
y
0
y
=
f
0
(x)
f(x)
Proof.
|y| = |f(x)|
ln|y| = ln|f(x)|
y
0
y
=
f
0
(x)
f(x)
x .
13.4
13.4.1
Definition 13.4.1 (). 위의 P Q P 가갈
PQ P .
13.4.2
Definition 13.4.2 (concave up funcion). f(x) , 이의
이은 ,
x
1
6= x
2
, f(
x
1
+ x
2
2
) <
f(x
1
) + f(x
2
)
2
Definition 13.4.3 (concave down function). f(x) , 이의
이은 ,
x
1
6= x
2
, f(
x
1
+ x
2
2
) >
f(x
1
) + f(x
2
)
2
Definition 13.4.4 (concave up). Let f(x) be a differentiable function on an interval I.
We will say that the graph of f(x) is concave up on I f
0
(x) is increasing on I
f
0
(x) f
00
(x) 0.
Definition 13.4.5 (concave down). Let f (x) be a differentiable function on an interval I.
We will say that the graph of f(x) is concave down on I f
0
(x) is decreasing on I
f
0
(x) f
00
(x) 0.
Theorem 13.4.1 ( ).
1. f
0
(x) > 0 f(x)
2. f(x) f
0
(x) 0 f(x)
13.4 78
3. f(x) f
0
(x) 0
4. f
00
(x) > 0 f(x)
5. f
0
(x) f
00
(x) 0 (f
0
(x) ) f(x)
6. f(x) f
00
(x) 0
Definition 13.4.6 (). y = f(x) 위의 P (a, f(a)) x = a
P
y = f (x) .
P (a, f(a)) f
00
(x) f
00
(a) f
00
(a) = 0.
Theorem 13.4.2 ( ).
y = f(x) f
00
(a) = 0, x = a f
00
(x) (a, f(a))
y = f(x) . ( f
00
(a) = 0 니다. ,
)
Note ( ).
If a function has a second derivative, the value of the second derivative is either 0 or undefined at each
of that function’s inflection points.
14
14.1 정적
Definition 14.1.1 (정적). f(x) f(x) 정적 .
Z
f(x)dx
. ,
d
dx
Z
f(x)dx = f (x)
. f(x) 정적 지지 는다. , f(x) 정적 F (x)
, f(x) 임의의 정적 F (x) + C ,
Z
f(x)dx = F (x) + C (where C is a constant)
.
R
integral symbol, C (the constant of integration), f (x)
(the integrand), x (the integration variable) , f (x) 정적
f(x) .
Theorem 14.1.1 (정적 ).
(F (x))
0
= f(x)
Z
f(x) = F (x) + C
Proof.
(
Z
f(x)dx F (x))
0
= f(x) f(x) = 0
Z
f(x)dx F (x) = C
Z
f(x)dx = F (x) + C
Theorem 14.1.2 ( 실수, , 정적).
f(x), g(x) 정적
1.
R
kf(x)dx = k
R
f(x)dx (, k 0 )
2.
R
{f(x) ± g(x)}dx =
R
f(x)dx ±
R
g(x)dx
14.2 정적 80
14.1.1 Reference Sites
-COS
14.2 정적
Definition 14.2.1 (정적).
Z
b
a
f(x)dx = lim
n→∞
n
X
k=1
f(x
k
)∆
x
(where x =
b a
n
, x
k
= a + x · k)
Theorem 14.2.1 ( 정적 ).
lim
n→∞
n
X
k=1
f(a +
pk
n
) ·
q
n
= q
Z
1
0
f(a + px)dx =
q
p
Z
p
0
f(a + x)dx
Proof. You can use Definition of definite integral or
pk
n
= x and
p
n
= dx. k
n p k 1 0 .
Proof. ,
pk
n
= x
k
and
p
n
= x . 정적
.
Note ( 정적 ).
Z
b
a
f(x)dx =
Z
ba
0
f(x + a)dx
= (b a)
Z
1
0
f((b a)x + a)dx
Theorem 14.2.2 (부분 ).
Z
f(x)g
0
(x)dx = f (x)g(x)
Z
f
0
(x)g(x)dx
Proof.
Note (tabular integration). tabular integration 방법 .
youtube 료로 https://youtu.be/E8N1E5ZAiIU .
Example 14.2.1 (부분 ).
f(x), g(x) [a, b] f
0
(x) =
1
4
g(x) , 정적
R
b
a
f(x)g(x)dx
?
Solution.
2[{f(b)}
2
{f(a)}
2
]
14.3 정적 81
Note ( ).
항함
R
β
α
a(x α)(x β)(x γ)dx =
a
6
(β α)
3
(γ
α + β
2
)
14.3 정적
Note ( · · ).
, , 역연
음이 .
Figure 14.1:
Theorem 14.3.1 ( ).
P t l = l(t) , v(t) =
dl
dt
= l
0
(t)
.
1. P ( t)
l(t) = l(t
0
)(= l
0
) +
R
t
t
0
v(x)dx
2. P (t = a t = b)
l(b) l(a) =
R
b
a
v(t)dt
3. P (t = a t = b)
s =
R
b
a
|v(t)|dt
15
15.1
Note (Hakchin- 방법). .
Definition 15.1.1 (). 결과
Definition 15.1.2 ( ).
Theorem 15.1.1 ( ). A, B , A m
, B n A B m+n
Theorem 15.1.2 ( ). A, B A m, 각각
B n A, B mn
15.1.1
A B = φ , n(A B) = n(A) + n(B)
A B 6= φ , n(A B) = n(A) + n(B) n(A B)
15.1.2
n(A × B) = n(A) × n(B)
15.2
15.2.1
Definition 15.2.1 (
n
P
r
). .
n
P
r
, n r 렬로
Definition 15.2.2 (
n
P
r
). .
n
P
r
= n!/(n r)!
n
P
n
= n!
0! = 1
n
P
0
= 1
15.3 83
15.2.2
Definition 15.2.3 (). n r
렬로 n r
1
n
Π
r
.
Theorem 15.2.1 ( Theorem).
n
Π
r
= n
r
Proof. ...
15.3
15.3.1
15.3.2
15.3.3 =
수수-
(derangement)
subfactorial
15.4
Note. , .
15.4.1
Definition 15.4.1 (). n r r
각각 n r (Combination)
n
C
r
.
n
r
.
15.4.2
1.
n
C
k
=
n
P
k
k!
=
n!
(n k)!k!
2.
n
C
k
=
n
C
nk
(n k)
3.
n
C
0
= 1
Theorem 15.4.1 ( Theorem1).
1.
n
C
k
=
n1
C
k1
+
n1
C
k
2.
n
C
k
=
n
k
·
n1
C
k1
1
repeated permutation
15.4 84
3. k ·
n
C
k
= n ·
n1
C
k1
4.
n
C
k
=
n k + 1
k
·
n
C
k1
5.
n
C
k
=
n
n k
·
n1
C
k
6.
n
C
k1
=
k
n k + 1
·
n
C
k
7.
n1
C
k
=
n k
n
·
n
C
k
8.
n
P
k
= n ·
n1
P
k1
9.
n
P
k
= (n k + 1) ·
n
P
k1
10.
n
P
k
= k ·
n1
P
k1
+
n1
P
k
(where 1 k < n)
Definition 15.4.2 (). 음으 누는 ,
음을 .
15.4.3
Definition 15.4.3 (). n r
2
n
H
r
.
Theorem 15.4.2 ( Theorem).
n
H
r
=
n1+r
C
r
Proof. n r n 1 r 렬로
H(n, r) =
(n 1 + r)!
(n 1)!r!
=
(n 1 + r)!
(n 1 + r r)!r!
= C(n 1 + r, r)
Note ( ).
Π(n, m)
n
P
m
S(m, n) ·
n
P
n
n
P
n
n
C
m
n
H
m
Question 15.4.1. X = {1, 2, 3, 4} f : X X ?
f 응이.
X 임의의 x f(x) 6= x .
Solution.
2
repeated combination
15.4 85
1. n(Ω) =
4
P
4
= 4! = 24
2. E : x X, f(x) 6= x
E
c
: x X, f(x) = x
3. n(E
c
) =
4
C
1
· 2 +
4
C
2
· 1 +
4
C
3
· 0 +
4
C
4
· 1 = 8 + 6 + 0 + 1 = 15
4. n(E) = 24 15 = 9
Example 15.4.1 ( ).
1, 2, 3, 4 3 : Π(4, 3)
4 3 .
1, 2, 3, 4 3 : Π(4, 3)
4 3 .
1, 2, 3, 4 3 방법 : Π(4, 3)
4 3 .
1, 2, 3, 4 3 :
4
H
3
4 3 .
1, 2, 3, 4 3 :
4
H
3
4 3 .
1, 2, 3, 4 3 방법 :
4
H
3
4 3 .
link-[]
link-- golang .
Example 15.4.2 ( ).
15.5 86
Figure 15.1: with
15.5
Theorem 15.5.1 (). . n (a + b)
n
.
(a + b)
n
=
n
C
0
a
0
b
n
+
n
C
1
a
1
b
n1
+
n
C
2
a
2
b
n2
+ ... +
n
C
k
a
k
b
nk
+ ... +
n
C
n
a
n
b
0
(a + b)
n
=
n
X
k=0
C(n, k)a
k
b
nk
=
n
X
k=0
C(n, k)a
nk
b
k
Theorem 15.5.2 ( ). 리를 (1 + x)
n
(x + 1)
n
=
n
C
0
+
n
C
1
x
1
+
n
C
2
x
2
+ ··· +
n
C
k
x
k
+ ··· +
n
C
n
x
n
.
1. : 2
n
=
n
C
0
+
n
C
1
+
n
C
2
+ ··· +
n
C
n
2. : 0 =
n
C
0
n
C
1
+
n
C
2
+ ··· + (1)
n
n
C
n
15.6 87
3. : 2
n1
=
n
C
0
+
n
C
2
+
n
C
4
+ ··· +
n
C
le
=
n
C
1
+
n
C
3
+
n
C
5
+ ··· +
n
C
lo
:
4. : 2
2n
=
2n+1
C
0
+
2n+1
C
1
+
2n+1
C
2
+ ··· +
2n+1
C
n
=
2n+1
C
n+1
=
2n+1
C
n+2
+ ··· +
2n+1
C
2n
+
2n+1
C
2n+1
5. n · 2
n1
= C(n, 1) + 2 · C(n, 2) + 3 · C(n, 3) + ··· + n · C(n, n)
Proof. 5 .
(x + 1)
n
=
n
C
0
+
n
C
1
x
1
+
n
C
2
x
2
+ ··· +
n
C
k
x
k
+ ··· +
n
C
n
x
n
C(n, k) =
n
k
C(n 1, k 1) = C(n, k) =
n(n 1)
k(k 1)
C(n 2, k 2) k · C(n, k) =
n · C(n 1, k 1)
,
P
n
k=0
k·
n
C
k
=
P
n
k=1
k·
n
C
k
=
P
n
k=1
n·
n1
C
k1
= n(
n1
C
0
+
n1
C
1
+···+
n1
C
n1
) = n·2
n1
15.6
15.6.1
,
Ferrers diagram .
Definition 15.6.1 ( ). n
n = n
1
+ n
2
+ n
3
+ ··· + n
k
(n n
1
n
2
n
2
n
3
··· n
k
1)
n
1
, n
2
, n
3
, ··· , n
k
.
Definition 15.6.2 ( ). n k(1 k n) 할하
P (n, k) .
Theorem 15.6.1 ( 1).
1 < k < n P (n, k) = P (n k, 1) + P (n k, 2) + ··· + P (n k, k)
1 < k < n P (n, k) = P (n 1, k 1) + P (n k, k)
P (n, k) = 0 (k > n)
Theorem 15.6.2 ( 2).
P (n, 1) = 1
P (n, 2) = [
n
2
]
···
P (n, n) = 1
P (n, n 1) = 1 (n 2)
P (n, n 2) = 2 (n 4)
P (n, n 3) = 3 (n 6)
15.6 88
P (n, n 4) = 5 (n 8)
Definition 15.6.3 (Q(n, k)).
Q(n, k) P (n, k) n 할하
Definition 15.6.4 (R(n, k)).
R(n, k) n k 할하
Definition 15.6.5 ().
n k 할하 = n k 할하
15.6.2
Definition 15.6.6 ( ). n k(1 k n)
부분 누는 . , A k 부분
A
1
, A
2
, ··· , A
k
할하
A
i
A
j
= (i, j = 1, 2, ··· , k i 6= j) (15.1)
A
1
A
2
··· A
k
= A (15.2)
.
Definition 15.6.7 ( ). n k(1 k n) 부분 할하
방법 S(n, k) .
Note.
S(n, k) S (Stirling, J.: 1692 - 1770) 자이.
Theorem 15.6.3 ( ).
p, q, r , C(n, p) · C(n p, q) · C(r, r)
p, q, r , C(n, p) ·C(n p, q) · C(r, r) ·
1
2!
p, q, r , C(n, p) ·C(n p, q) · C(r, r) ·
1
3!
Theorem 15.6.4 ( S(n, k) ).
1 < k < n S(n, k) = S(n 1, k 1) + S(n 1, k) · k
Proof. n {1, 2, 3, ··· , n} k 부분 할하 방법 위의
.
n 부분 ( 임의의 부분
.)
n 부분 {1, 2, 3, ··· , n 1} k 1 할하 {n}
함하 k . k 할하
S(n 1, k 1)
15.6 89
n 1 부분
부분 {1, 2, 3, ··· , n 1} k 할한 , k 부분
n k . k 할하
S(n 1, k) ·k
k n 부분 .
음이 .
S(n, k) = S(n 1, k 1) + S(n 1, k) · k
Corollary 15.6.4.1 ( S(n, k) ).
S(n, k) = S(n 1, k 1) + S(n 1, k) ·k = S(n 1, k 1) ·k
0
+ S(n 1 1, k 1) ·k
1
+ S(n 1
2, k 1) · k
2
+ ··· + S(n 1 (n k), k 1) ·k
nk
Theorem 15.6.5 ( S(n, k) ).
1. S(n, 1) = 1
2. S(n, 2) =
2
n
2
2
= 2
n1
1
3. S(n, n 1) =
n
C
2
4. S(n, n) = 1
5. S(n, k) = S(n 1, k 1) + S(n 1, k) · k
Proof. 할할 U = {1, 2, ··· , n} .
1. 1 부분 할하 방법 {1, 2, ··· , n} .
2. .
Figure 15.2:
15.6 90
.
S(17, 2) 17 2 부분 할하
S(17, 2) = C(17, 16) · C(1, 1) + C(17, 15) · C(2, 2) + C(17, 14) · C(3, 3) + ··· + C(17, 9) · C(8, 8)
= C(17, 1) + C(17, 2) + C(17, 3) + ··· + C(17, 8)
= (C(17, 0) + C(17, 1) + C(17, 2) + C(17, 3) + ··· + C(17, 8)) C(17, 0)
= 2
16
1
3. A = {a
1
, a
2
, ··· , a
n
} {a
1
, a
2
}{a
3
}···{a
n
}, {a
1
, a
3
}{a
2
}···{a
n
}, ···
. , n 2 부분 , (n-2)
1 부분 . n 2
방법 S(n, n 1) =
n
C
2
.
4. n 부분 할하 방법 {1} {2} {3} ··· {n} .
5. n 1 부분 n 1 부분
.
Note ( ).
: 5 3 담는
S(5, 3) · 3! = 25 · 6 = 150
: 5 3 담는
3
Π
5
= 3
5
= 243
: 5 3 담는
3
H
2
=
4
C
2
= 6
: 5 3 담는
3
H
5
=
7
C
5
= 21
: 5 3 담는
P (5, 3) = 2
: 5 3 담는
P (5, 1) + P (5, 2) + P (5, 3) = 1 + 2 + 2 = 5
: 5 3 담는
S(5, 3) = S(4, 2) + S(4, 3) · 3 = 7 + 6 · 3 = 7 + 18 = 25
: 5 3 담는
S(5, 1) + S(5, 2) + S(5, 3) = 1 + 15 + 25 = 41
16
16.1
16.1.1
16.1.2
16.1.3
16.2
16.2.1
Definition 16.2.1 ( - the conditional probability of A given B).
B 다는 A B A
. P (A|B) . , .
P (A|B) =
P (A B)
P (B)
where P (B) > 0
Definition 16.2.2 ( - the conditional probability of A given B).
A B P (A|B) . , .
P (A|B) =
P (A B)
P (B)
where P (B) > 0
16.2.2
Definition 16.2.3 ( ).
A B P (A|B) = P (A)
A B P (A|B) 6= P (A)
Theorem 16.2.1 ( ).
A B P (A B) = P (A)P (B)(if P (A) > 0, P (B) > 0)
Theorem 16.2.2 (Bayes’ Theorem). 우와 relative weight (
) 적절 .
16.2 92
Table 16.1: Visualization of Bayes’ Theorem
B B
c
A P (B) · P (A|B) P (B
c
) · P (A|B
c
)
A
c
P (B) · P (A
c
|B) P (B
c
) · P (A
c
|B
c
)
16.2.3
Definition 16.2.4 (). 우와 반복 ,
결과가 결과 ,
.
Theorem 16.2.3 ( ). 1 E p , n
E r P
r
P
r
=
n
C
r
p
r
(1 p)
nr
(r = 0, 1, 2, ..., n)
17
Note (Reference Sites).
discrete random variable - 부분
17.1
Definition 17.1.1 (). , ,
Definition 17.1.2 (). 간격
Definition 17.1.3 ( ). ,
Definition 17.1.4 ( ).
Definition 17.1.5 ().
Example 17.1.1 (). a b :
a + b
2
= ( )/2
Definition 17.1.6 ().
Definition 17.1.7 (). ,
Definition 17.1.8 ().
Example 17.1.2 (). , , , ...
Definition 17.1.9 (). = ( )/( )
Note (). = ()*() /()
Definition 17.1.10 (). 앙에
Example 17.1.3 ( ). 작은 례로
1. N (N+1)/2
2. N N/2 (N/2 +1)
Note ( ). .
Definition 17.1.11 ().
Note (). .
17.2 94
17.2
17.2.1
Definition 17.2.1 (-). 결과 공간 실수
, 각각 .
X, Y, Z, ··· , x, y, z, ··· . X
x P (X = x) .
Note (). 공간 실수
.
Note ().
probability space = (Ω, F, P )
P (X = x) = P ({ω |X(ω) = x}) where X : E(measurable space) and P : F
[0, 1] and F is a set of events
P (X S) = P ({ω |X(ω) S}) where P is the probability measure on (Ω, F) and S E
Note ().
F = P(Ω)
Note ().
P (X = x) =
n(X = x)
n(Ω)
=
n({ω |X(ω) = x})
n(Ω)
17.2.2
Definition 17.2.2 (-). X
있을 , X .
Example 17.2.1 (). 1
X X 0, 1, 2, 3 X .
Definition 17.2.3 (). A random variable X : E is a measurable function from a
set of possible outcomes to a measurable space E. The technical axiomatic definition requires to
be a probability space (see Measure-theoretic definition). Usually X is real-valued (i.e. E = R).
Definition 17.2.4 ().
X : Ω E(measurable space) Img(X) is countable(= finite denumerable).
If the image is not countable then X is called a continuous random variable.
Definition 17.2.5 (probability mass function). X x
1
, x
1
, x
2
, . . . , x
n
, X p
1
, p
1
, p
2
, . . . , p
n
X f
X
(x
i
) =
P (X = x
i
) = p
i
(i = 1, 2, . . . , n) X .
Note ( X ). X
.
17.2 95
Figure 17.1:
Theorem 17.2.1 ( ). X P (X = x
i
) = p
i
(i =
1, 2, 3, y, n) , 음이 .
0 f
X
(x) 1
P
xImg(X)
f
X
(x) = 1
P (x
i
X x
j
) =
P
j
k=i
p
k
(where i j i, j = 1, 2, ··· , n)
Theorem 17.2.2 ( ).
E(X) = x
1
p
1
+ x
2
p
2
+ ··· + x
n
p
n
=
n
X
i=1
x
i
p
i
Definition 17.2.6 ( ).
V (X) = E((X m)
2
) =
n
X
i=i
(x
i
m)
2
p
i
Theorem 17.2.3 ( ).
V (X) = E(X
2
) {E(X)}
2
Proof.
V (X) = E((X m)
2
) =
n
X
i=i
(x
i
m)
2
p
i
=
n
X
i=i
(x
2
i
p
i
2mx
i
p
i
+ m
2
p
i
)
=
n
X
i=i
x
2
i
p
i
2m
n
X
i=i
x
i
p
i
+ m
2
n
X
i=i
p
i
= E(X
2
) {E(X)}
2
Theorem 17.2.4 ( aX + b , , ).
E(aX + b) = aE(X) + b
V (aX + b) = a
2
V (X)
σ(aX + b) = |a|σ(X)
17.2 96
Definition 17.2.7 (). 1 E p , n
E X X
f
X
(k) = P (X = k) =
n
C
k
p
k
q
nk
(k = 0, 1, 2, . . . , n, q = 1 p)
. , B(n, p) .
Note ( 해해 ).
Independent Bernoulli Trial (or Binomial Trial)
1. X :
2. A : , (IBT), 추출
3. A
P (A) :
, A (IBT Event)
4. : n
5. X = occurrence of A (A )
f
X
(k) = P (X = k) =
n
C
k
p
k
q
nk
(k = 0, 1, 2, . . . , n, q = 1 p)
X B(n, p) (where p = P (A))
6. X .
X = x 0 1 2 ··· x ··· n
P (X = x)
n
C
0
p
0
q
n
n
C
1
p
1
q
n1
n
C
2
p
2
q
n2
···
n
C
x
p
k
q
nx
···
n
C
n
p
n
q
0
1
7. 위의 (p + q)
n
.
(p + q)
n
=
n
C
0
p
0
q
n
+
n
C
1
p
1
q
n1
+
n
C
2
p
2
q
n2
+ ... +
n
C
x
p
x
q
nx
+ ... +
n
C
n
p
n
q
0
(p + q)
n
=
n
X
x=0
C(n, x)p
x
q
nx
=
n
X
x=0
C(n, x)p
nx
q
x
Theorem 17.2.5 ( Bn, p , , ).
E(X) = np
V (X) = npq
σ(X) =
npq
Proof. ....
V (X) = E((X m)
2
) =
n
X
i=i
(x
i
m)
2
p
i
17.2 97
=
n
X
i=i
(x
2
i
p
i
2mx
i
p
i
+ m
2
p
i
)
=
n
X
i=i
x
2
i
p
i
2m
n
X
i=i
x
i
p
i
+ m
2
n
X
i=i
p
i
= E(X
2
) {E(X)}
2
Theorem 17.2.6 (LLN:Law of Large Numbers). 1 A
p , n A X . 작은 h
n P (|
X
n
p| < h) 1 . .
Note (LLN). n 대도,
X
n
p 점점 .
.
17.2.3
Definition 17.2.8 (-). X 실수 ,
X .
Example 17.2.2 ( ).
OX 3 각각 임의 , X X 0, 1, 2, 3
있으 .
5 AB 위의 C 임의 잡을 , AC X X
0 < X < 5 실수 있으 .
Figure 17.2:
Definition 17.2.9 (-pdf). α X β 실수 X
α X β f(x) , f(x) X
.
f
X
(x) 0
y = f (x) x x = α, x = β 1.
Z
β
α
f(x)dx = 1
17.2 98
P (a X b) y = f (x) x x = a, x = b
. (, α a b β)
P (a X b) =
Z
b
a
f(x)dx
Definition 17.2.10 (). X f(x) f(x) =
1
2πσ
e
1
2
·(
x m
σ
)
2
(where x R, m = constant, σ > 0, e = 2.71828 ···) , X ,
N(m, σ
2
) . , f(x) . X
N(m, σ
2
) .
Theorem 17.2.7 ( ).
1. f(x) x 이의 1.
2. x=m
1
2πσ
3. x=m , x.
4. σ , m .
Figure 17.3: σ
5. m , σ .
Figure 17.4: m
Definition 17.2.11 (). N(0, 1)
Theorem 17.2.8 ( pdf). Z , Z
17.3 99
f(z) =
1
2π
e
1
2
·z
2
(where < z < )
Theorem 17.2.9 ( ). X N(m, σ
2
) ,
Z =
X m
σ
N(0, 1) .
P (a X b) = P (
a m
σ
Z
b m
σ
)
Proof.
E(Z) = E(
X m
σ
) =
1
σ
(E(X) m) = 0
V (Z) = V (
X m
σ
) =
1
σ
2
V (X) = 1
17.2.4
Definition 17.2.12 (ˆn 1).
n . n 1 np 5 nq 5
ˆn . ˆn 1 ˆn 30
Theorem 17.2.10 ( 이의 ).
X B(n, p) n 1 X ≈∼ N (np, npq) (where q = 1 p)
17.3
17.3.1
Definition 17.3.1 (). 방법
Example 17.3.1.
Definition 17.3.2 ().
방법
Example 17.3.2. ,
Definition 17.3.3 ().
Definition 17.3.4 (). , 추출
.
Example 17.3.3. 생산 알아 생산
1000 임의
17.3 100
: 생산
: 임의 1000
: 1000
Definition 17.3.5 (임의추출). 추출 추출 방법
Note. 임의추출 추출 .
Note (추출 추출 1).
1. 추출: 추출 추출
2. 추출: 추출 추출
Note (추출 추출 2).
임의추출 , ,
.
추출 추출 .
Note (추출 추출 - ).
N n 임의추출
추출 :
N
Π
n
= N
n
추출 :
1 추출 V
N
P
n
추출 V
N
C
n
Definition 17.3.6 (Random sampling). Random sampling is a part of the sampling technique in
which each sample has an equal probability of being chosen. A sample chosen randomly is meant to
be an unbiased representation of the total population.
Note (Hakchin Thought).
ˆ
X
1
,
ˆ
X
2
, ··· ,
ˆ
X
i
, ··· .
Note (Hakchin Thought). , 임의 추출 음을 추출 . ( :
n, : ˆn)
E(
ˆ
X
i
) =
n
X
k=1
( ˆx
i
)
k
· P ((
ˆ
X
i
) = ( ˆx
i
)
k
) = E(X) = m
where
¯
X =
1
ˆn
(
ˆ
X
1
+
ˆ
X
2
+ ··· +
ˆ
X
ˆn
) =
1
ˆn
P
ˆn
i=1
ˆ
X
i
and m is a mean of a population
Definition 17.3.7 (임의 - Random sample Unbiased sample). 임의추출 임의
.
Definition 17.3.8 ( - Biased sample). 측치
얻어 .
Definition 17.3.9 (추출 ).
17.3 101
추출 : 추출 방법
추출 : 추출 방법
17.3.2
Definition 17.3.10 (, , ).
X
X : symbol : m
X : symbol : σ
2
X : symbol : σ
Definition 17.3.11 (, , ). ˆn x
1
, x
2
, ··· , x
ˆn
임의 추출
추출 : (Sample mean) symbol :
¯
X
추출 : symbol : S
2
(Bessel’s correction )
추출 : symbol : S
1.
¯
X =
1
ˆn
(X
1
+ X
2
+ ··· + X
ˆn
) =
1
ˆn
P
ˆn
i=1
X
i
2. S
2
=
1
ˆn 1
{(X
1
¯
X)
2
+ (X
2
¯
X)
2
+ ··· + (X
ˆn
¯
X)
2
} =
1
ˆn 1
P
ˆn
i=1
(X
i
¯
X)
2
3. S =
S
2
=
r
1
ˆn 1
P
ˆn
i=1
(X
i
¯
X)
2
Note (
¯
X, S
2
, S).
¯
X, S
2
, S 각각 .
Definition 17.3.12 (a random sample of size n).
The random variables
ˆ
X
i
constitute a random sample of size n if and only if:
1. the
ˆ
X
i
are independent, and
2. the
ˆ
X
i
are identically distributed, that is, each
ˆ
X
i
comes from the same distribution f(x) with
mean m and variance σ
2
.
3. (i.e. E(X
i
) = m, V (X
i
) = σ
2
)
We say that the
ˆ
X
i
are i.i.d. (The first i. stands for independent, and the i.d. stands for identically
distributed.)
Note (Hakchin Thought). , 임의 추출 음을 추출 .
E(X
i
) = m, V (X
i
) = σ
2
17.3 102
Theorem 17.3.1 ( , , ). m, σ
2
ˆn X
1
, X
2
, ··· , X
ˆn
임의추출 ,
¯
X , , .
¯
X : : E(
¯
X) = m
¯
X : : V (
¯
X) =
σ
2
ˆn
¯
X : : σ(
¯
X) =
σ
ˆn
Proof. Starting with the definition of the sample mean, we have:
E(
¯
X) = E(
1
ˆn
(X
1
+ X
2
+ ··· + X
ˆn
))
Then, using the linear operator property of expectation, we get:
E(
¯
X) =
1
ˆn
(E(X
1
) + E(X
2
) + ··· + E(X
ˆn
))
Now, the X
k
are identically distributed, which means they have the same mean m. Therefore, replacing
E(X
k
) with the alternative notation m, we get:
E(
¯
X) =
1
ˆn
(m + m + ··· + m)
Now, because there are ˆn m’s in the above formula, we can rewrite the expected value as:
E(
¯
X) =
1
ˆn
(ˆnm) = m
We have shown that the mean (or expected value, if you prefer) of the sample mean
¯
X is m. That is,
we have shown that the mean of
¯
X is the same as the mean of the individual X
k
.
Proof. Starting with the definition of the sample mean, we have:
V (
¯
X) = V (
1
ˆn
(X
1
+ X
2
+ ··· + X
ˆn
))
Rewriting the term on the right so that it is clear that we have a linear combination of X
k
’s, we get:
V (
¯
X) = V (
1
ˆn
X
1
+
1
ˆn
X
2
+ ··· +
1
ˆn
X
ˆn
)
Then, applying the theorem on a variance, we get:
V (
¯
X) =
1
ˆn
2
V (X
1
) +
1
ˆn
2
V (X
2
) + ··· +
1
ˆn
2
V (X
ˆn
)
Now, the X
k
are identically distributed, which means they have the same variance σ
2
. Therefore,
replacing V (X
k
) with the alternative notation σ
2
, we get:
V (
¯
X) =
1
ˆn
2
(σ
2
+ σ
2
+ ··· + σ
2
)
Now, because there are ˆn σ
2
’s in the above formula, we can rewrite the variance of
¯
X as:
V (
¯
X) =
1
ˆn
2
(ˆ
2
) =
σ
2
ˆn
17.3 103
Theorem 17.3.2 ( ). Let = ˆn
X N(m, σ
2
)
¯
X N(m, (σ(
¯
X))
2
) where (σ(
¯
X))
2
=
σ
2
ˆn
(17.1)
(¬(X N(m, σ
2
) (X N (m, σ
2
)) (ˆn 1)
¯
X ≈∼ N(m,
σ
2
ˆn
) (17.2)
Definition 17.3.13 (()).
() = |
¯
X m|
Definition 17.3.14 ( , , ).
: 추출 등등
, ( )
( )
(Confidence Level) : ,
(Confidence Interval) :
Theorem 17.3.3 (CI(m) : ( α%)).
¯
X k
σ
ˆn
m
¯
X + k
σ
ˆn
Or
CI(m) = [
¯
X kσ(
¯
X),
¯
X + kσ(
¯
X)]
Proof.
P (k Z
¯
X
k) =
α
100
P (k
¯
X m
σ
ˆn
k) =
α
100
P (
¯
X k
σ
ˆn
m
¯
X + k
σ
ˆn
) =
α
100
Theorem 17.3.4 ( ( 95%)).
¯
X 1.96
σ
ˆn
m
¯
X + 1.96
σ
ˆn
Proof.
P (k Z
¯
X
k) =
α
100
P (k Z
¯
X
k) =
95
100
17.3 104
P (1.96 Z
¯
X
1.96) =
95
100
P (1.96
¯
X m
σ
ˆn
1.96) =
95
100
P (
¯
X 1.96
σ
ˆn
m
¯
X + 1.96
σ
ˆn
) =
95
100
Theorem 17.3.5 ( ( 99%)).
¯
X 2.58
σ
ˆn
m
¯
X + 2.58
σ
ˆn
Proof.
P (k Z
¯
X
k) =
α
100
P (k Z
¯
X
k) =
99
100
P (2.58 Z
¯
X
2.58) =
99
100
P (2.58
¯
X m
σ
ˆn
2.58) =
99
100
P (
¯
X 2.58
σ
ˆn
m
¯
X + 2.58
σ
ˆn
) =
99
100
Note ( : S ).
σ 부분. ˆn (ˆn 30),
S s σ .
17.3.3 율의 율의
율의
Definition 17.3.15 ( p).
X = ()
p =
X
n
Definition 17.3.16 ( ˆp).
ˆ
X =
, ˆn
ˆ
X , 율은
.
ˆp =
ˆ
X/ˆn
17.3 105
Theorem 17.3.6. 율이 p ˆn 임의추출 ,
ˆ
X B(ˆn, p)
ˆ
X B(ˆn, p)
Note (
ˆ
X, ˆp).
ˆ
X 추출 ,
ˆp .
Theorem 17.3.7 (율의 , , ).
E(ˆp) = E(
ˆ
X
ˆn
) =
1
ˆn
· E(
ˆ
X) =
1
ˆn
· ˆnp = p
V (ˆp) = V (
ˆ
X
ˆn
) =
1
ˆn
2
· V (
ˆ
X) =
1
ˆn
2
· ˆnpq =
pq
ˆn
σ(ˆp) =
p
V (ˆp) =
r
pq
ˆn
Proof.
ˆ
X B(ˆn, p)
Theorem 17.3.8. If ˆn 1,
ˆ
X ≈∼ N(ˆnp, ˆnpq) (17.3)
ˆ
X ≈∼ N(ˆnp, ˆnˆpˆq) (17.4)
Theorem 17.3.9 ( - ). p, 임의추출 ˆn,
ˆp, ˆn 1 ˆp ≈∼ N (p,
pq
ˆn
), Z
ˆp
=
ˆp p
r
pq
ˆn
≈∼ N (0, 1) (where q = 1 p)
Proof.
E(Z) = E(
ˆp p
r
pq
ˆn
) =
s
ˆn
pq
(E(ˆp) p) = 0
V (Z) = V (
ˆp p
r
pq
ˆn
) =
ˆn
pq
(V (ˆp)) = 1
Note (ˆn 1).
ˆn 다는 ˆnp 5, ˆnq 5 .
Corollary 17.3.9.1 ( - ). p, 임의추출
ˆn, ˆp, ˆn 1 ˆp ≈∼ N (p,
ˆpˆq
ˆn
), Z
ˆp
=
ˆp p
r
ˆpˆq
ˆn
≈∼ N (0, 1) (where ˆq = 1 ˆp)
17.3 106
율의
Definition 17.3.17 (()).
() = |ˆp p|
Theorem 17.3.10 (CI(p) : 율의 ( α% )).
ˆp k
r
ˆpˆq
ˆn
p ˆp + k
r
ˆpˆq
ˆn
Proof.
P (k Z
ˆp
k) =
α
100
P (k
ˆp p
r
ˆpˆq
ˆn
k) =
α
100
P (ˆp k
r
ˆpˆq
ˆn
p ˆp + k
r
ˆpˆq
ˆn
) =
α
100
Theorem 17.3.11 (율의 ( 95% )).
ˆp 1.96
r
ˆpˆq
ˆn
p ˆp + 1.96
r
ˆpˆq
ˆn
Proof.
P (k Z
ˆp
k) =
α
100
P (k Z
ˆp
k) =
95
100
P (1.96 Z
ˆp
1.96) =
95
100
P (1.96
ˆp p
r
ˆpˆq
ˆn
1.96) =
95
100
P (ˆp 1.96
r
ˆpˆq
ˆn
p ˆp + 1.96
r
ˆpˆq
ˆn
) =
95
100
Theorem 17.3.12 (율의 ( 99% )).
ˆp 2.58
r
ˆpˆq
ˆn
p ˆp + 2.58
r
ˆpˆq
ˆn
17.3 107
Proof.
P (k Z
ˆp
k) =
α
100
P (k Z
ˆp
k) =
99
100
P (2.58 Z
ˆp
2.58) =
99
100
P (2.58
ˆp p
r
ˆpˆq
ˆn
2.58) =
99
100
P (ˆp 2.58
r
ˆpˆq
ˆn
p ˆp + 2.58
r
ˆpˆq
ˆn
) =
99
100
18
Note ().
: 0 90 작은
: 90
: 90 180 작은
: 180
: , 2
: , 2 작은
: 90 ,
: 180 ,
Definition 18.0.1 ( ). 공간 행하
.
18.1
3:4:5
5:12:13
7:24:25
8:15:17
18.2
Theorem 18.2.1 ( ).
Theorem 18.2.2 ( ).
ABCD , 이의 이의 곱과 ,
.
AB · CD + BC · DA = AC · BD
Theorem 18.2.3 ( ).
접점
.
18.3 Hilbert’s axioms 109
18.3 Hilbert’s axioms
Hilbert’s axioms
18.4
Definition 18.4.1 ( 23).
1. 부분 .
2. 이이.
3. .
4. 위의 .
5. .
6. .
7. 위의 .
8. 있으 이의
.
9. , .
10. 있을 , .
11. ( , ) .
12. ( ) 작은 .
13. (경계) .
14. () .
15. 위의 있으 같게
.
16. .
17. . .
18. . .
19. . .
20. , ,
.
21. . .
.
22. 같고 .
. 마름 . 행한 .
.
18.4 110
23. 있으
19
Definition 19.0.1 ().
위의 P (x, y) P
0
(x
0
, y
0
)
f : (x, y) (x
0
, y
0
)
, f .
Definition 19.0.2 ( ). sdfdsafsad
Theorem 19.0.1 ( ). f 2 × 1 X
1
, X
2
1. f (X
1
+ X
2
) = f(X
1
) + f(X
2
)
2. f (kX
1
) = kf(X
1
)
3. f (kX
1
+ hX
2
) = kf(X
1
) + hf(X
2
) where k, h R
Theorem 19.0.2 (). fsadfsda
20
20.1
Definition 20.1.1 (2 - ). 실수 x y
, , Ax
2
+By
2
+Cxy +Dx+Ey +F = 0
. xy Ax
2
+ By
2
+ Cx + Dy + E = 0 .
20.1.1
Definition 20.1.2 (, a parabola). 위의 정점 F F l
, F l
Definition 20.1.3 (, a parabola). a parabola P = {P |P F = P l} where F : a fixed point, the
focus, l: a fixed line, the directrix
Definition 20.1.4 ( ). a vertex, an axis of symmetry
Theorem 20.1.1 ( ). F(f, 0) x=-f
y
2
= 4fx (where f 6= 0)
f .
Figure 20.1: Parabola
Proof.
Note ( ).
Ax
2
+ By
2
+ Cx + Dy + E = 0 where (A = B 6= 0) (C
2
+ D
2
4AE > 0)
20.1 113
(
C
2A
,
D
2A
)
C
2
+ D
2
4AE
2A
Note ( ).
Ax
2
+ By
2
+ Cx + Dy + E = 0 (where (A = 0 BC 6= 0) (B = 0 AD 6= 0))
20.1.2
Definition 20.1.5 (, an ellipse). 위의 정점 F
1
, F
2
. F
1
, F
2
.
Definition 20.1.6 (, an ellipse). an ellipse E = {P |P F
1
+ P F
2
= 2v} where F
1
, F
2
: fixed points,
the foci v: 2v > F
1
F
2
> 0
Definition 20.1.7 ( ). a vertex, , ,
Theorem 20.1.2 ( ). F
1
(f, 0), F
2
(f, 0) 2v
x
2
v
2
+
y
2
b
2
= 1 (where v > b > 0, f
2
+ b
2
= v
2
)
.
Figure 20.2: Ellipse
Proof.
Note ( ).
Ax
2
+ By
2
+ Cx + Dy + E = 0 (where AB > 0, A 6= B)
20.1.3
Definition 20.1.8 (, a hyperbola). 위의 정점 F
1
, F
2
. F
1
, F
2
.
Definition 20.1.9 (, an hyperbola). a hyperbola H = {P ||P F
1
P F
2
| = 2v} where F
1
, F
2
:
fixed points, the foci v: v > 0
20.1 114
Definition 20.1.10 ( ). a vertex, , ,
Theorem 20.1.3 ( ). F
1
(f, 0), F
2
(f, 0) 2v
x
2
v
2
+
y
2
b
2
= 1 (where f > v > 0, v
2
+ b
2
= f
2
)
.
Figure 20.3: Hyperbola
Proof.
Theorem 20.1.4 ( ). F
1
(f, 0), F
2
(f, 0) 2v
x
2
v
2
+
y
2
b
2
= 1 (where f > v > 0, b
2
= f
2
v
2
)
.
y = ±
b
v
x
Theorem 20.1.5 ( ).
20.1 115
Figure 20.4: Hyperbola2
Proof.
1. 리를 .
2. .
3. LEF T
test
RIGHT
20.2 116
Theorem 20.1.6 ( ). F
1
(0, f), F
2
(0, f) 2v
x
2
a
2
+
y
2
v
2
= 1 (where v > a > 0, a
2
= f
2
v
2
)
.
y = ±
v
a
x
Note ( ).
Ax
2
+ By
2
+ Cx + Dy + E = 0 (where AB < 0)
20.2
20.2.1
Definition 20.2.1. f(x, y) = 0 , x y 조정
y x . y x .
20.2.2
20.2.3
Theorem 20.2.1 ( m ).
1. y
2
= 4fx
y = mx +
f
m
2. x
2
= 4fy
y = mx +
f
m
(m)
3
= mx fm
2
Theorem 20.2.2 ( m ).
1.
x
2
v
2
+
y
2
b
2
= 1
y = mx ±
m
2
v
2
+ b
2
2.
x
2
a
2
+
y
2
v
2
= 1
y = mx ±
m
2
a
2
+ v
2
Theorem 20.2.3 ( m ).
1.
x
2
v
2
+
y
2
b
2
= 1
20.2 117
y = mx ±
m
2
v
2
b
2
2.
x
2
a
2
+
y
2
v
2
= 1
y = mx ±
m
2
a
2
+ v
2
21 공간
Definition 21.0.1 (공간 ).
Definition 21.0.2 (공간 ).
Definition 21.0.3 ( 공간 , ).
Definition 21.0.4 (공간 ).
Note (
P A ·
P B ).
1. AB M
P A ·
P B = |
P M|
2
|
MB|
2
2.
P A ·
P B = 0 P AB 위의
21.0.1
Theorem 21.0.1 ( 행한 ).
A ~u 행한 l 위의 P .
1. vector equation : A, P 각각 ~a, ~p , A ~u 행한 l
~p = ~a + t~u (where t R)
~u = (u
1
, u
2
, u
3
) l .
2. parametric equation : x = x
1
+ u
1
t, y = y
1
+ u
2
t, z = z
1
+ u
3
t
3. Symmetry equation : P P(x, y, z) , A(x
1
, y
1
, z
1
) ~u =
(u
1
, u
2
, u
3
) 행한 l x = x
1
+ u
1
t, y = y
1
+ u
2
t, z = z
1
+ u
3
t,
t .
A(x
1
, y
1
, z
1
) ~u = (a, b, c) 행한 l
x x
1
u
1
=
y y
1
u
2
=
z z
1
u
3
(where abc 6= 0)
Definition 21.0.5 ( ).
21.0.2
Theorem 21.0.2 ( ).
A ~n α 위의 P .
1. A, P 각각 ~a, ~p , A ~n α
(~p ~a) ·~n = 0
21 공간 119
~n α , a normal vector .
2. P P(x, y, z) , A(x
1
, y
1
, z
1
) ~n = (n
1
, n
2
, n
3
)
α n
1
(x x
1
) + n
2
(y y
1
) + n
3
(z z
1
) = 0
Note ( ).
n
1
x + n
2
y + n
3
z + d = 0 (where n
1
, n
2
, n
3
0 )
Definition 21.0.6 ( ).
Definition 21.0.7 ( 이의 ).
Definition 21.0.8 ( ).
Question 21.0.1.
Solution.
22 공간
22.1
Note (공간 ). , ,
Theorem 22.1.1 (공간 ). , , , ,
. 공간 공간 . 공간
, , 이의 .
1. .
2. 위의 .
3. .
Figure 22.1: 공간
위의 .
Theorem 22.1.2 ( ).
1.
2. 행하
Figure 22.2:
Theorem 22.1.3 ( ).
1.
2.
3.
4. 행한
22.1 121
Figure 22.3:
Proof.
1. A, B, C α .
공간
. A, B, C α .
Figure 22.4: blur blur
2. l A α .
l 위의 임의의 B, C A, B, C
α . l위의 B, C α 있으
l α . l A α .
Figure 22.5: blur blur
3. A l, m α .
A l 위의 임의의 B , m 위의 임의의 C
A, B, C α . l 위의
A, B α 있으 l α , m 위의 A, C α
있으 m α . A l, m α
.
Figure 22.6: blur blur
4. 행한 l, m α .
m 위의 임의의 A A l l
A α . l, m 함하 l m 위의 A
함하 α . 행한 l, m α .
22.2 122
Figure 22.7: blur blur
Example 22.1.1 ( ). 음은 ABCD .
A, B, C
B CD
AD CD
AB CD
Figure 22.8: example
Definition 22.1.1 (). .
22.2
Theorem 22.2.1 ( ).
.
.
는다.
행하.
. (skew)
Theorem 22.2.2 ( ).
.
.
.
는다.
행하.
Theorem 22.2.3 ( ).
22.3 123
.
.
는다.
행하.
22.3
22.3.1
Lemma 22.3.0.1 ( ).
Figure 22.9:
Theorem 22.3.1 ( 0).
1. (α k β) (l = α γ) (m = β γ) (l k m)
(α k β) (α γ) k (β γ)
2. (l k α) (l β) (m = α β) (m k l)
(l k α) (l β) (l k α β)
3. (l k α) (l β) (l γ) (m = α β) (n = α γ) (m k n)
4. (α k β) (l α) (l k β)
5. (l k m) (l α) (m * α) (α k m)
6. (P / α) ({P } = l m) (l, m k α) (5lm k α)
7. (α k β) (β k γ) (α k γ)
22.3 124
8. (a k c) (b k c) (a k b)
Theorem 22.3.2 ( 1).
1. (l k m) (l α) (m * α) (α k m)
Figure 22.10:
proof ¬(α k m) {P } = α m P l = α 5lm P l m ( l k m)
2. (l k α) (l β) (m = α β) (m k l)
Figure 22.11:
proof l k α l α = ( m α) (l k m skew(l, m)) l k m ( l, m β)
3. (P / α) ({P } = l m) (l, m k α) (5lm k α)
Figure 22.12:
proof ¬(5lm k α) n = 5lm α n α l k n n k m l k m ( {P } = l m)
4. (α k β) (l = α γ) (m = β γ) (l k m)
Figure 22.13:
proof α k β α β = l m = ( l α m β) l k m skew(l, m) l k m ( l, m γ)
22.3 125
5. (l k α) (l β) (l γ) (m = α β) (n = α γ) (m k n)
Figure 22.14:
proof l k α l m = l k m ( l, m β) m k γ ( l γ) m n = ( n γ) m k n (
m, n α)
6. (α k β) (β k γ) (α k γ)
Figure 22.15:
proof ¬(α k γ) (l, m α) ({P } = l m) l k β m k β ( α k β) l γ 6= m γ 6= (
¬(α k γ)) l β 6= m β 6= ( β k γ) ( l k β m k β) α k γ ( l, m α)
7. (a k c) (b k c) (a k b)
Figure 22.16:
proof . a k c
a, c α . , a 위의 A b
β α β a
0
a
0
k c. a, a
0
A
c 행하 . , a
0
k b a k b .
Proof. visang p166
Theorem 22.3.3 ( 2).
1. l m 행할 , l 함하 m 함하 α m 행하.
2. l α 행할 , l 함하 β α m l 행하.
3. α P l, m α 행하 l, m
β α 행하.
22.3 126
4. 행한 α, β γ 각각 l, m , l, m
행하.
5. l α 행하 l 함하 β, γ α 각각 m, n
, m, n 행하.
6. α, β, γ α k β, β k γ α k γ.
7. l, m, n l k m, m k n l k n.
Definition 22.3.1 ( ).
ssen page 178
l, m l m
l
0
l
0
, m . l
0
m
l, m . l, m 90 l, m
, l m .
Figure 22.17:
Definition 22.3.2 ( ).
ssen page 178
l α O , l위의 A α
H AOH l α .
Figure 22.18:
Definition 22.3.3 ( ).
l α 위의 , l α . l α
. l α , l α O
.
Theorem 22.3.4 ( ). l α O O
α 위의 m, n 각각 l α .
Proof. visang p167
22.4 127
Figure 22.19:
22.4
Theorem 22.4.1 (theorem of three perpendiculars - Hakchin).
A
B C
l
α
Figure 22.20:
(l a) (α b) (c l)
(α b) (c l) (l a)
(c l) (l a) (a b) (α b)
Proof. 이의 , 공간
. 리를 .
22.4 128
1. 1) (l α) (l b)
2) (l 5ABC) (c l)
2. 1) (l α) (l b)
2) (l 5ABC) (l a)
3. 1) (c l) (l a) (l 5ABC)
2) (l b) (a b) (α b)
Theorem 22.4.2 (
1
).
:
(l AO) (α OP ) (P A l)
(α OP ) (P A l) (l AO)
(P A l) (l AO) (AO OP ) (α OP )
Figure 22.21:
Proof. 이의 , 공간
. 리를 .
1. (l AO) (α OP )
(l α) (l OP )
(l 5P AO) (P A l)
2. (α OP ) (P A l)
(l α) (l OP )
(l 5P AO) (l AO)
3. (P A l l AO l 5P AO) l OP
(AO OP ) (OP α위의 AO l α OP )
1
theorem of three perpendiculars
22.5 129
22.5
Definition 22.5.1 (). 위의 부분 누는 각각 부분
.
Definition 22.5.2 (). l α, β
. l , α, β 각각 .
l 위의 O l OA, OB α, β 각각 ,
AOB O . ( p174)
.
Figure 22.22: dihedral angle
Definition 22.5.3 ( ).
.
Note ( ). 작은 .
Note (&). , 공간
.
Theorem 22.5.1 ( ).
1. l α l β β α
l α l 함하 β α .
Figure 22.23:
2. α β A β O = F ootp(A, α β) AO α
α β A β (A α, β O) AO α
α, β , β위의 A α, β
O AO α .
22.6 , orthograph 130
Figure 22.24:
3. α β α γ l = β γ l α
α β, γ l l α .
Figure 22.25:
Proof. visang p175, ssen p186
22.6 , orthograph
Definition 22.6.1 (). α A α A
0
, A
0
A α . F α
F
0
F α .
Note (). () .
Definition 22.6.2 ( - ).
l α , l α l
0
l l
0
l α .
Note ( ). ,
.
Theorem 22.6.1 ( ). AB α A
0
B
0
, AB
α θ
A
0
B
0
= ABcosθ
Figure 22.26:
22.6 , orthograph 131
Proof.
Figure 22.27:
Theorem 22.6.2 ( ). β S, α
S
0
, α, β θ(0 θ
π
2
)
S
0
= Scosθ
Figure 22.28:
Proof.
22.7 공간 132
Figure 22.29:
22.7 공간
22.7.1 Reference Sites
Octant (solid geometry)
Note ( ).
x
2
+ y
2
+ z
2
+ Ax + By + Cz + D = 0 where (A
2
+ B
2
+ C
2
4D > 0)
(
A
2
,
B
2
,
C
2
)
A
2
+ B
2
+ C
2
4D
2
23
23.1
23.2
23.3
Theorem 23.3.1 ( ).
P t x x = f(t) , t P v(t)
a(t) .
1. v(t) =
dx
dt
= f
0
(t)
2. a(t) =
d
dt
v(t) = f
00
(t)
Theorem 23.3.2 ( ).
위의 P (x, y) t x = f(t), y = g(t) .
1. P (x, y)
v = (v
x
, v
y
) = (
dx
dt
,
dy
dt
) = (f
0
(t), g
0
(t))
2. P (x, y)
a = (a
x
, a
y
) = (
d
2
x
dt
2
,
d
2
y
dt
2
) = (f
00
(t), g
00
(t))
Theorem 23.3.3 ( - ).
P t x = f(t) , v(t) =
dx
dt
= f
0
(t)
.
1. P ( t)
f(t) = f(t
0
)(= x
0
) +
R
t
t
0
v(x)dx
2. P () (t = a t = b)
f(b) f(a) =
R
b
a
v(t)dt
3. P () (t = a t = b)
s =
R
b
a
|v(t)|dt
Theorem 23.3.4 ( - ).
위의 P (x, y) t x = f(t), y = g(t) ,
p
.
23.3 134
1. P ( t)
p = (x, y) = (f(t), g(t)) = (f (t
0
), g(t
0
)) + (
R
t
t
0
f
0
(x)dx,
R
t
t
0
g
0
(y)dy)
2. P (t = a t = b)
(f(b), g(b)) (f(a), g(a)) = (
R
b
a
f
0
(t)dt,
R
b
a
g
0
(t)dt)
3. P (t = a t = b)
s =
R
b
a
|
v |dt =
R
b
a
|(f
0
(t), g
0
(t))|dt
Partl [3] has proposed that ...
Bibliography
[1] ...
[2] ...
[3] H. Partl: German T
E
X, TUGboat Volume 9, Issue 1 (1988)
Index
, 85